Let $p$ be an odd prime, and let $p-1$ be a multiple of 4. Prove $mathbb F_p^{times}$ contains an element of...












1












$begingroup$


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?




  1. 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 $x^2-1$).


  2. 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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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?




  1. 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 $x^2-1$).


  2. 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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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?




  1. 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 $x^2-1$).


  2. 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







share|cite|improve this question











$endgroup$




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?




  1. 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 $x^2-1$).


  2. 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 1:32







Ekhin Taylor R. Wilson

















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


















  • $begingroup$
    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










1 Answer
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$begingroup$

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 $4$.






share|cite|improve this answer









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    $begingroup$

    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 $4$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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 $4$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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 $4$.






        share|cite|improve this answer









        $endgroup$



        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 $4$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 2:31









        moutheticsmouthetics

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