$f: G to mathbb{C}^*$ is a homomorphism. Show that the sum $sum f (g) = 0$ or $n$












5












$begingroup$


Let $ mathbb{C}^*$ be the multiplicative group of non-zero complex numbers. Let $G$ be an abelian group and suppose $f: G to mathbb{C}^*$ is a homomorphism. Prove that $sum_{g in G} f(g)=n$ or, $sum_{g in G} f(g)=0$, where $n =o(G)$



Proof attempt:



The case is evident for the trivial homomorphism; the sum adds up to $n$.



For the second part



We know, the only elements with finite order in the group $ mathbb{C}^*$ are $1$ and $-1$, with $o(-1)=2$.



Now, the only case when $f(g)$ can take $-1$ as a value is when $n$ is even.



Consider the subgroup $({1, -1}, .) = G'$ of the group $ mathbb{C}^*$. We have, from the Isomorphism Theorem, $ G/ ker( f ) simeq G' $ [since $f$ takes each value from $G'$].



As $o(G')=2$, $o(G/ ker( f ))=2$, i.e $o(ker (f))= n/2$. Hence, when summed, the resultant is $0$.



Edit: A foolish assumption has been taken. The finite ordered complex numbers in the said group is of the form $z^n=1$, so I have 'proved' a very restricted case, which is not at all desired.






abstract-algebra group-theory proof-verification finite-groups group-isomorphism







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    What's the order of $i$ then?
    $endgroup$
    – the_fox
    Jan 4 at 3:11






  • 2




    $begingroup$
    Does G have to have finite order?
    $endgroup$
    – Joel Pereira
    Jan 4 at 3:39










  • $begingroup$
    @JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
    $endgroup$
    – Robert Lewis
    Jan 4 at 3:45










  • $begingroup$
    @the_fox :( back to square one.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:41






  • 1




    $begingroup$
    @RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
    $endgroup$
    – Joel Pereira
    Jan 4 at 15:33
















5












$begingroup$


Let $ mathbb{C}^*$ be the multiplicative group of non-zero complex numbers. Let $G$ be an abelian group and suppose $f: G to mathbb{C}^*$ is a homomorphism. Prove that $sum_{g in G} f(g)=n$ or, $sum_{g in G} f(g)=0$, where $n =o(G)$



Proof attempt:



The case is evident for the trivial homomorphism; the sum adds up to $n$.



For the second part



We know, the only elements with finite order in the group $ mathbb{C}^*$ are $1$ and $-1$, with $o(-1)=2$.



Now, the only case when $f(g)$ can take $-1$ as a value is when $n$ is even.



Consider the subgroup $({1, -1}, .) = G'$ of the group $ mathbb{C}^*$. We have, from the Isomorphism Theorem, $ G/ ker( f ) simeq G' $ [since $f$ takes each value from $G'$].



As $o(G')=2$, $o(G/ ker( f ))=2$, i.e $o(ker (f))= n/2$. Hence, when summed, the resultant is $0$.



Edit: A foolish assumption has been taken. The finite ordered complex numbers in the said group is of the form $z^n=1$, so I have 'proved' a very restricted case, which is not at all desired.






abstract-algebra group-theory proof-verification finite-groups group-isomorphism







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    What's the order of $i$ then?
    $endgroup$
    – the_fox
    Jan 4 at 3:11






  • 2




    $begingroup$
    Does G have to have finite order?
    $endgroup$
    – Joel Pereira
    Jan 4 at 3:39










  • $begingroup$
    @JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
    $endgroup$
    – Robert Lewis
    Jan 4 at 3:45










  • $begingroup$
    @the_fox :( back to square one.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:41






  • 1




    $begingroup$
    @RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
    $endgroup$
    – Joel Pereira
    Jan 4 at 15:33














5












5








5


1



$begingroup$


Let $ mathbb{C}^*$ be the multiplicative group of non-zero complex numbers. Let $G$ be an abelian group and suppose $f: G to mathbb{C}^*$ is a homomorphism. Prove that $sum_{g in G} f(g)=n$ or, $sum_{g in G} f(g)=0$, where $n =o(G)$



Proof attempt:



The case is evident for the trivial homomorphism; the sum adds up to $n$.



For the second part



We know, the only elements with finite order in the group $ mathbb{C}^*$ are $1$ and $-1$, with $o(-1)=2$.



Now, the only case when $f(g)$ can take $-1$ as a value is when $n$ is even.



Consider the subgroup $({1, -1}, .) = G'$ of the group $ mathbb{C}^*$. We have, from the Isomorphism Theorem, $ G/ ker( f ) simeq G' $ [since $f$ takes each value from $G'$].



As $o(G')=2$, $o(G/ ker( f ))=2$, i.e $o(ker (f))= n/2$. Hence, when summed, the resultant is $0$.



Edit: A foolish assumption has been taken. The finite ordered complex numbers in the said group is of the form $z^n=1$, so I have 'proved' a very restricted case, which is not at all desired.






abstract-algebra group-theory proof-verification finite-groups group-isomorphism







share|cite|improve this question











$endgroup$




Let $ mathbb{C}^*$ be the multiplicative group of non-zero complex numbers. Let $G$ be an abelian group and suppose $f: G to mathbb{C}^*$ is a homomorphism. Prove that $sum_{g in G} f(g)=n$ or, $sum_{g in G} f(g)=0$, where $n =o(G)$



Proof attempt:



The case is evident for the trivial homomorphism; the sum adds up to $n$.



For the second part



We know, the only elements with finite order in the group $ mathbb{C}^*$ are $1$ and $-1$, with $o(-1)=2$.



Now, the only case when $f(g)$ can take $-1$ as a value is when $n$ is even.



Consider the subgroup $({1, -1}, .) = G'$ of the group $ mathbb{C}^*$. We have, from the Isomorphism Theorem, $ G/ ker( f ) simeq G' $ [since $f$ takes each value from $G'$].



As $o(G')=2$, $o(G/ ker( f ))=2$, i.e $o(ker (f))= n/2$. Hence, when summed, the resultant is $0$.



Edit: A foolish assumption has been taken. The finite ordered complex numbers in the said group is of the form $z^n=1$, so I have 'proved' a very restricted case, which is not at all desired.






abstract-algebra group-theory proof-verification finite-groups group-isomorphism




abstract-algebra group-theory proof-verification finite-groups group-isomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 11:44







Subhasis Biswas

















asked Jan 4 at 2:51









Subhasis BiswasSubhasis Biswas

512311




512311








  • 6




    $begingroup$
    What's the order of $i$ then?
    $endgroup$
    – the_fox
    Jan 4 at 3:11






  • 2




    $begingroup$
    Does G have to have finite order?
    $endgroup$
    – Joel Pereira
    Jan 4 at 3:39










  • $begingroup$
    @JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
    $endgroup$
    – Robert Lewis
    Jan 4 at 3:45










  • $begingroup$
    @the_fox :( back to square one.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:41






  • 1




    $begingroup$
    @RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
    $endgroup$
    – Joel Pereira
    Jan 4 at 15:33














  • 6




    $begingroup$
    What's the order of $i$ then?
    $endgroup$
    – the_fox
    Jan 4 at 3:11






  • 2




    $begingroup$
    Does G have to have finite order?
    $endgroup$
    – Joel Pereira
    Jan 4 at 3:39










  • $begingroup$
    @JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
    $endgroup$
    – Robert Lewis
    Jan 4 at 3:45










  • $begingroup$
    @the_fox :( back to square one.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:41






  • 1




    $begingroup$
    @RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
    $endgroup$
    – Joel Pereira
    Jan 4 at 15:33








6




6




$begingroup$
What's the order of $i$ then?
$endgroup$
– the_fox
Jan 4 at 3:11




$begingroup$
What's the order of $i$ then?
$endgroup$
– the_fox
Jan 4 at 3:11




2




2




$begingroup$
Does G have to have finite order?
$endgroup$
– Joel Pereira
Jan 4 at 3:39




$begingroup$
Does G have to have finite order?
$endgroup$
– Joel Pereira
Jan 4 at 3:39












$begingroup$
@JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
$endgroup$
– Robert Lewis
Jan 4 at 3:45




$begingroup$
@JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
$endgroup$
– Robert Lewis
Jan 4 at 3:45












$begingroup$
@the_fox :( back to square one.
$endgroup$
– Subhasis Biswas
Jan 4 at 11:41




$begingroup$
@the_fox :( back to square one.
$endgroup$
– Subhasis Biswas
Jan 4 at 11:41




1




1




$begingroup$
@RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
$endgroup$
– Joel Pereira
Jan 4 at 15:33




$begingroup$
@RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
$endgroup$
– Joel Pereira
Jan 4 at 15:33










2 Answers
2






active

oldest

votes


















6












$begingroup$

It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @Shaun: nice edit, thanks!
    $endgroup$
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    $begingroup$
    Amazing. Just amazing.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40










  • $begingroup$
    Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    $endgroup$
    – Cameron Williams
    Jan 4 at 12:03










  • $begingroup$
    @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 5 at 0:41



















0












$begingroup$

Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you please verify mine?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 6:00










  • $begingroup$
    the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    $endgroup$
    – Riley
    Jan 4 at 7:01












  • $begingroup$
    $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 8:27










  • $begingroup$
    I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    $endgroup$
    – Riley
    Jan 4 at 10:37










  • $begingroup$
    I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061265%2ff-g-to-mathbbc-is-a-homomorphism-show-that-the-sum-sum-f-g-0-or%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @Shaun: nice edit, thanks!
    $endgroup$
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    $begingroup$
    Amazing. Just amazing.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40










  • $begingroup$
    Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    $endgroup$
    – Cameron Williams
    Jan 4 at 12:03










  • $begingroup$
    @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 5 at 0:41
















6












$begingroup$

It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @Shaun: nice edit, thanks!
    $endgroup$
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    $begingroup$
    Amazing. Just amazing.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40










  • $begingroup$
    Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    $endgroup$
    – Cameron Williams
    Jan 4 at 12:03










  • $begingroup$
    @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 5 at 0:41














6












6








6





$begingroup$

It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.






share|cite|improve this answer











$endgroup$



It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 6:13









Shaun

9,406113684




9,406113684










answered Jan 4 at 3:58









Robert LewisRobert Lewis

48k23067




48k23067








  • 1




    $begingroup$
    @Shaun: nice edit, thanks!
    $endgroup$
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    $begingroup$
    Amazing. Just amazing.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40










  • $begingroup$
    Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    $endgroup$
    – Cameron Williams
    Jan 4 at 12:03










  • $begingroup$
    @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 5 at 0:41














  • 1




    $begingroup$
    @Shaun: nice edit, thanks!
    $endgroup$
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    $begingroup$
    Amazing. Just amazing.
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40










  • $begingroup$
    Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    $endgroup$
    – Cameron Williams
    Jan 4 at 12:03










  • $begingroup$
    @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 5 at 0:41








1




1




$begingroup$
@Shaun: nice edit, thanks!
$endgroup$
– Robert Lewis
Jan 4 at 6:23




$begingroup$
@Shaun: nice edit, thanks!
$endgroup$
– Robert Lewis
Jan 4 at 6:23




1




1




$begingroup$
Amazing. Just amazing.
$endgroup$
– Subhasis Biswas
Jan 4 at 11:40




$begingroup$
Amazing. Just amazing.
$endgroup$
– Subhasis Biswas
Jan 4 at 11:40












$begingroup$
Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
$endgroup$
– Cameron Williams
Jan 4 at 12:03




$begingroup$
Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
$endgroup$
– Cameron Williams
Jan 4 at 12:03












$begingroup$
@SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
$endgroup$
– Robert Lewis
Jan 5 at 0:41




$begingroup$
@SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
$endgroup$
– Robert Lewis
Jan 5 at 0:41











0












$begingroup$

Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you please verify mine?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 6:00










  • $begingroup$
    the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    $endgroup$
    – Riley
    Jan 4 at 7:01












  • $begingroup$
    $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 8:27










  • $begingroup$
    I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    $endgroup$
    – Riley
    Jan 4 at 10:37










  • $begingroup$
    I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40
















0












$begingroup$

Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you please verify mine?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 6:00










  • $begingroup$
    the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    $endgroup$
    – Riley
    Jan 4 at 7:01












  • $begingroup$
    $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 8:27










  • $begingroup$
    I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    $endgroup$
    – Riley
    Jan 4 at 10:37










  • $begingroup$
    I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40














0












0








0





$begingroup$

Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.






share|cite|improve this answer









$endgroup$



Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 4:16









RileyRiley

1825




1825












  • $begingroup$
    Can you please verify mine?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 6:00










  • $begingroup$
    the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    $endgroup$
    – Riley
    Jan 4 at 7:01












  • $begingroup$
    $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 8:27










  • $begingroup$
    I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    $endgroup$
    – Riley
    Jan 4 at 10:37










  • $begingroup$
    I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40


















  • $begingroup$
    Can you please verify mine?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 6:00










  • $begingroup$
    the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    $endgroup$
    – Riley
    Jan 4 at 7:01












  • $begingroup$
    $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 8:27










  • $begingroup$
    I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    $endgroup$
    – Riley
    Jan 4 at 10:37










  • $begingroup$
    I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    $endgroup$
    – Subhasis Biswas
    Jan 4 at 11:40
















$begingroup$
Can you please verify mine?
$endgroup$
– Subhasis Biswas
Jan 4 at 6:00




$begingroup$
Can you please verify mine?
$endgroup$
– Subhasis Biswas
Jan 4 at 6:00












$begingroup$
the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
$endgroup$
– Riley
Jan 4 at 7:01






$begingroup$
the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
$endgroup$
– Riley
Jan 4 at 7:01














$begingroup$
$z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
$endgroup$
– Subhasis Biswas
Jan 4 at 8:27




$begingroup$
$z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
$endgroup$
– Subhasis Biswas
Jan 4 at 8:27












$begingroup$
I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
$endgroup$
– Riley
Jan 4 at 10:37




$begingroup$
I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
$endgroup$
– Riley
Jan 4 at 10:37












$begingroup$
I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
$endgroup$
– Subhasis Biswas
Jan 4 at 11:40




$begingroup$
I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
$endgroup$
– Subhasis Biswas
Jan 4 at 11:40


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061265%2ff-g-to-mathbbc-is-a-homomorphism-show-that-the-sum-sum-f-g-0-or%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more