Let $p$ be an odd prime, and let $p-1$ be a multiple of 4. Prove $mathbb F_p^{times}$ contains an element of...
$begingroup$
Let $G=mathbb F_p^{times}$. Consider the homomorphism $varphi: G to G$ defined by $varphi(x)=x^2$. The image of $varphi$ has order $frac{p-1}{2}$. This is an even integer because $frac{p-1}{2} = 2 frac{p-1}{4}$, and $frac{p-1}{4}$ is an integer by assumption. Since groups that have even order have an element of order 2, the image of $varphi$ contains an element of order 2, call it $y$. $y$ has a preimage $x$ where $x^2=y$. Since $y$ has order 2, $x$ has order 4.
My question: Am I implicitly using any of these facts?
The kernel is ${1,-1}$ or equivalently ${1,p-1}$ (because a polynomial with degree $2$, in this case $x^2-1$ has at most $2$ zeros, and ${1,-1}$ are zeros of $x^2-1$).
First Sylow theorem
I think the kernel is used to compute the order of the image, but I think I don't need the First Sylow theorem because I use the weaker fact that a group of even order has an element of order 2.
Thanks in advance!
abstract-algebra
asked Jan 3 at 0:43
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
$endgroup$
|
show 9 more comments
$begingroup$
Let $G=mathbb F_p^{times}$. Consider the homomorphism $varphi: G to G$ defined by $varphi(x)=x^2$. The image of $varphi$ has order $frac{p-1}{2}$. This is an even integer because $frac{p-1}{2} = 2 frac{p-1}{4}$, and $frac{p-1}{4}$ is an integer by assumption. Since groups that have even order have an element of order 2, the image of $varphi$ contains an element of order 2, call it $y$. $y$ has a preimage $x$ where $x^2=y$. Since $y$ has order 2, $x$ has order 4.
My question: Am I implicitly using any of these facts?
The kernel is ${1,-1}$ or equivalently ${1,p-1}$ (because a polynomial with degree $2$, in this case $x^2-1$ has at most $2$ zeros, and ${1,-1}$ are zeros of $x^2-1$).
First Sylow theorem
I think the kernel is used to compute the order of the image, but I think I don't need the First Sylow theorem because I use the weaker fact that a group of even order has an element of order 2.
Thanks in advance!
abstract-algebra
asked Jan 3 at 0:43
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
$endgroup$
$begingroup$
The group $;G;;$ is of order $;p-1=0pmod 4;$ and since thus it has one unique subgroup of any order dividing its order, it also has an element of order $;4;$ (in fact, exactly two elements of order four)
$endgroup$
– DonAntonio
Jan 3 at 0:46
$begingroup$
@DonAntonio I don't think I understand. If $G$ has order $n$, and $m$ divides $n$, $G$ doesn't necessarily have a subgroup of order $m$ right?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 0:52
$begingroup$
You're using the first result to show that $#operatorname{im} varphi = frac{1}{2} #G$. (That wouldn't be the case, if for example, $G = mathbb{Z}_2times mathbb{Z}_2$; but $G$ is cyclic.)
$endgroup$
– anomaly
Jan 3 at 1:01
$begingroup$
@anomaly If $G$ is cyclic of order $2n$, then $G^2$ has order $n$?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 1:12
$begingroup$
@anomaly What is the exact relationship between $G$ is cyclic and kernel is ${1,-1}$?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 1:14
|
show 9 more comments
$begingroup$
Let $G=mathbb F_p^{times}$. Consider the homomorphism $varphi: G to G$ defined by $varphi(x)=x^2$. The image of $varphi$ has order $frac{p-1}{2}$. This is an even integer because $frac{p-1}{2} = 2 frac{p-1}{4}$, and $frac{p-1}{4}$ is an integer by assumption. Since groups that have even order have an element of order 2, the image of $varphi$ contains an element of order 2, call it $y$. $y$ has a preimage $x$ where $x^2=y$. Since $y$ has order 2, $x$ has order 4.
My question: Am I implicitly using any of these facts?
The kernel is ${1,-1}$ or equivalently ${1,p-1}$ (because a polynomial with degree $2$, in this case $x^2-1$ has at most $2$ zeros, and ${1,-1}$ are zeros of $x^2-1$).
First Sylow theorem
I think the kernel is used to compute the order of the image, but I think I don't need the First Sylow theorem because I use the weaker fact that a group of even order has an element of order 2.
Thanks in advance!
abstract-algebra
asked Jan 3 at 0:43
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
$endgroup$
Let $G=mathbb F_p^{times}$. Consider the homomorphism $varphi: G to G$ defined by $varphi(x)=x^2$. The image of $varphi$ has order $frac{p-1}{2}$. This is an even integer because $frac{p-1}{2} = 2 frac{p-1}{4}$, and $frac{p-1}{4}$ is an integer by assumption. Since groups that have even order have an element of order 2, the image of $varphi$ contains an element of order 2, call it $y$. $y$ has a preimage $x$ where $x^2=y$. Since $y$ has order 2, $x$ has order 4.
My question: Am I implicitly using any of these facts?
The kernel is ${1,-1}$ or equivalently ${1,p-1}$ (because a polynomial with degree $2$, in this case $x^2-1$ has at most $2$ zeros, and ${1,-1}$ are zeros of $x^2-1$).
First Sylow theorem
I think the kernel is used to compute the order of the image, but I think I don't need the First Sylow theorem because I use the weaker fact that a group of even order has an element of order 2.
Thanks in advance!
abstract-algebra
abstract-algebra
asked Jan 3 at 0:43
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
asked Jan 3 at 0:43
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
edited Jan 3 at 1:32
Ekhin Taylor R. Wilson
asked Jan 3 at 0:43
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
asked Jan 3 at 0:43
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
asked Jan 3 at 0:43
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
558
$begingroup$
The group $;G;;$ is of order $;p-1=0pmod 4;$ and since thus it has one unique subgroup of any order dividing its order, it also has an element of order $;4;$ (in fact, exactly two elements of order four)
$endgroup$
– DonAntonio
Jan 3 at 0:46
$begingroup$
@DonAntonio I don't think I understand. If $G$ has order $n$, and $m$ divides $n$, $G$ doesn't necessarily have a subgroup of order $m$ right?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 0:52
$begingroup$
You're using the first result to show that $#operatorname{im} varphi = frac{1}{2} #G$. (That wouldn't be the case, if for example, $G = mathbb{Z}_2times mathbb{Z}_2$; but $G$ is cyclic.)
$endgroup$
– anomaly
Jan 3 at 1:01
$begingroup$
@anomaly If $G$ is cyclic of order $2n$, then $G^2$ has order $n$?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 1:12
$begingroup$
@anomaly What is the exact relationship between $G$ is cyclic and kernel is ${1,-1}$?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 1:14
|
show 9 more comments
$begingroup$
The group $;G;;$ is of order $;p-1=0pmod 4;$ and since thus it has one unique subgroup of any order dividing its order, it also has an element of order $;4;$ (in fact, exactly two elements of order four)
$endgroup$
– DonAntonio
Jan 3 at 0:46
$begingroup$
@DonAntonio I don't think I understand. If $G$ has order $n$, and $m$ divides $n$, $G$ doesn't necessarily have a subgroup of order $m$ right?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 0:52
$begingroup$
You're using the first result to show that $#operatorname{im} varphi = frac{1}{2} #G$. (That wouldn't be the case, if for example, $G = mathbb{Z}_2times mathbb{Z}_2$; but $G$ is cyclic.)
$endgroup$
– anomaly
Jan 3 at 1:01
$begingroup$
@anomaly If $G$ is cyclic of order $2n$, then $G^2$ has order $n$?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 1:12
$begingroup$
@anomaly What is the exact relationship between $G$ is cyclic and kernel is ${1,-1}$?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 1:14
$begingroup$
The group $;G;;$ is of order $;p-1=0pmod 4;$ and since thus it has one unique subgroup of any order dividing its order, it also has an element of order $;4;$ (in fact, exactly two elements of order four)
$endgroup$
– DonAntonio
Jan 3 at 0:46
$begingroup$
The group $;G;;$ is of order $;p-1=0pmod 4;$ and since thus it has one unique subgroup of any order dividing its order, it also has an element of order $;4;$ (in fact, exactly two elements of order four)
$endgroup$
– DonAntonio
Jan 3 at 0:46
$begingroup$
@DonAntonio I don't think I understand. If $G$ has order $n$, and $m$ divides $n$, $G$ doesn't necessarily have a subgroup of order $m$ right?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 0:52
$begingroup$
@DonAntonio I don't think I understand. If $G$ has order $n$, and $m$ divides $n$, $G$ doesn't necessarily have a subgroup of order $m$ right?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 0:52
$begingroup$
You're using the first result to show that $#operatorname{im} varphi = frac{1}{2} #G$. (That wouldn't be the case, if for example, $G = mathbb{Z}_2times mathbb{Z}_2$; but $G$ is cyclic.)
$endgroup$
– anomaly
Jan 3 at 1:01
$begingroup$
You're using the first result to show that $#operatorname{im} varphi = frac{1}{2} #G$. (That wouldn't be the case, if for example, $G = mathbb{Z}_2times mathbb{Z}_2$; but $G$ is cyclic.)
$endgroup$
– anomaly
Jan 3 at 1:01
$begingroup$
@anomaly If $G$ is cyclic of order $2n$, then $G^2$ has order $n$?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 1:12
$begingroup$
@anomaly If $G$ is cyclic of order $2n$, then $G^2$ has order $n$?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 1:12
$begingroup$
@anomaly What is the exact relationship between $G$ is cyclic and kernel is ${1,-1}$?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 1:14
$begingroup$
@anomaly What is the exact relationship between $G$ is cyclic and kernel is ${1,-1}$?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 1:14
|
show 9 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Note that we have by Wilson theorem $$(p-1)! equiv -1 mod p$$
Then
begin{align}-1& =(p-1)!\
& = prod_{k=1}^{p-1 over 2} k(p-k)\
& = (-1)^{p-1 over 2}prod_{k=1}^{p-1 over 2}k^2\
&= a^2 qquad (text{ because } pequiv 1 mod 4)end{align}
Hence $a^2=-1$ and so $a $ has order $4$.
answered Jan 3 at 2:31
moutheticsmouthetics
50937
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060151%2flet-p-be-an-odd-prime-and-let-p-1-be-a-multiple-of-4-prove-mathbb-f-p%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that we have by Wilson theorem $$(p-1)! equiv -1 mod p$$
Then
begin{align}-1& =(p-1)!\
& = prod_{k=1}^{p-1 over 2} k(p-k)\
& = (-1)^{p-1 over 2}prod_{k=1}^{p-1 over 2}k^2\
&= a^2 qquad (text{ because } pequiv 1 mod 4)end{align}
Hence $a^2=-1$ and so $a $ has order $4$.
answered Jan 3 at 2:31
moutheticsmouthetics
50937
$endgroup$
add a comment |
$begingroup$
Note that we have by Wilson theorem $$(p-1)! equiv -1 mod p$$
Then
begin{align}-1& =(p-1)!\
& = prod_{k=1}^{p-1 over 2} k(p-k)\
& = (-1)^{p-1 over 2}prod_{k=1}^{p-1 over 2}k^2\
&= a^2 qquad (text{ because } pequiv 1 mod 4)end{align}
Hence $a^2=-1$ and so $a $ has order $4$.
answered Jan 3 at 2:31
moutheticsmouthetics
50937
$endgroup$
add a comment |
$begingroup$
Note that we have by Wilson theorem $$(p-1)! equiv -1 mod p$$
Then
begin{align}-1& =(p-1)!\
& = prod_{k=1}^{p-1 over 2} k(p-k)\
& = (-1)^{p-1 over 2}prod_{k=1}^{p-1 over 2}k^2\
&= a^2 qquad (text{ because } pequiv 1 mod 4)end{align}
Hence $a^2=-1$ and so $a $ has order $4$.
answered Jan 3 at 2:31
moutheticsmouthetics
50937
$endgroup$
Note that we have by Wilson theorem $$(p-1)! equiv -1 mod p$$
Then
begin{align}-1& =(p-1)!\
& = prod_{k=1}^{p-1 over 2} k(p-k)\
& = (-1)^{p-1 over 2}prod_{k=1}^{p-1 over 2}k^2\
&= a^2 qquad (text{ because } pequiv 1 mod 4)end{align}
Hence $a^2=-1$ and so $a $ has order $4$.
answered Jan 3 at 2:31
moutheticsmouthetics
50937
answered Jan 3 at 2:31
moutheticsmouthetics
50937
answered Jan 3 at 2:31
moutheticsmouthetics
50937
answered Jan 3 at 2:31
moutheticsmouthetics
50937
50937
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060151%2flet-p-be-an-odd-prime-and-let-p-1-be-a-multiple-of-4-prove-mathbb-f-p%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
The group $;G;;$ is of order $;p-1=0pmod 4;$ and since thus it has one unique subgroup of any order dividing its order, it also has an element of order $;4;$ (in fact, exactly two elements of order four)
$endgroup$
– DonAntonio
Jan 3 at 0:46
$begingroup$
@DonAntonio I don't think I understand. If $G$ has order $n$, and $m$ divides $n$, $G$ doesn't necessarily have a subgroup of order $m$ right?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 0:52
$begingroup$
You're using the first result to show that $#operatorname{im} varphi = frac{1}{2} #G$. (That wouldn't be the case, if for example, $G = mathbb{Z}_2times mathbb{Z}_2$; but $G$ is cyclic.)
$endgroup$
– anomaly
Jan 3 at 1:01
$begingroup$
@anomaly If $G$ is cyclic of order $2n$, then $G^2$ has order $n$?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 1:12
$begingroup$
@anomaly What is the exact relationship between $G$ is cyclic and kernel is ${1,-1}$?
$endgroup$
– Ekhin Taylor R. Wilson
Jan 3 at 1:14