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Let $g \in G$ and $g$ has odd order $n$. I have already shown that there is a unique square root in $H=<g>$ by showing an isomorphism from H to H as $\phi(h)=h^2$. I now have to show that if $x$ is a square root of $g$ and $x$ is not in H=$<g>$ then $|x|=2n$.

I also have to show two other parts.

  1. If $|G|$ is odd then every element in G has a unique square root.

  2. If $|g|=m$ with m being even there there is no element in $H=<g>$ with $x^2=g$. This square root could lie outside of H. Show that $(13)(24) \in S_4$ does have a square root.

Could I get some tips on how to go about doing this perhaps?

Jack Armstrong
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  • For (1), see http://math.stackexchange.com/q/522273/589. – lhf Nov 09 '14 at 23:05
  • I suppose with |x| you mean the order of x. Now, if x is a square root of g, the orders of x and g are related. How? – jflipp Nov 09 '14 at 23:12
  • Regarding (2), assume we have x in H with $x^2 = g$. Since $x \in H$, we have $x = g^k$ for some $k$. So, we have $x^2 = g^{2k} = g$. This implies that $2k \equiv 1 \mod 1$. Why? Is this possible? – jflipp Nov 09 '14 at 23:16

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Take the group $\langle x \rangle$, $H$ is subgroup of $\langle x \rangle$ and since both of them are finite you can apply Lagrange theorem: $|\langle x \rangle| = |H| * [\langle x \rangle : H] = 2n$. (You know that $[\langle x \rangle : H] = 2$ since $x^2 = g \in H$)

If $|G| = 2n+1$ take $a \in G$, $\langle a^{-1} \rangle$ is a subgroup of $G$ and therefore $|\langle a^{-1} \rangle|$ divides $|G|$. So $(a^{-1})^{|G|} = e \Rightarrow (a^{-1})^{2n+1} = e \Rightarrow (a^{-1})^{2n} = a \Rightarrow (a^{-n})^2 = a$. So square roots exist. If $x^2 = y^2$, then $x = xe = xx^{2n+1} = x^{2n+2} = (x^2)^{n+1} = (y^2)^{n+1} = y^{2n+1}y = ey = y$. This shows the uniqueness.

And for the last part if $x = g^n \in \langle g \rangle$ is such that $x^2 = g$ then $g^{2n} = g \Rightarrow g^{2n+m-1} = g^m = e \Rightarrow m$ divides $2n+m-1$ which is impossible since $m$ is even. Which is the desired contradiction.

brick
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  • how is H a subgroup of ? Is it just that is in G? – Jack Armstrong Nov 09 '14 at 23:12
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    @Jack $g = x^2 \in $ therefore the whole $H$ is subset of $$, since $g$ generates it. – brick Nov 09 '14 at 23:14
  • how is $(a^{-1})^{2n}=a$, which then implies that $(a^{-n})^2=a$? Also for the last part, how did you get $g^{2n}=g$ to imply that $g^{2n-1+m}=g^m=e$? and how do you come to the conclusion that m then divides 2n-1+m? – Jack Armstrong Nov 10 '14 at 00:38
  • @Jack I multiplied both sides of $(a^{-1})^{2n+1} = e$ by $a$. That's how. In finite groups if you take an element $g$ to the power of the number of elements of the group you get $e$ (the neutral element). And I multiplied $g^{n2} = g$ by $g^{m-1}$ to get the next equality. And if you have $g^t = e$ this means that the order of $g$ divides $t$. Pff, long question :D – brick Nov 10 '14 at 00:48
  • when we use lagrane's theorem, that gives us the fact that 2n is the smallest right? – Jack Armstrong Nov 10 '14 at 17:08
  • @Jack The smallest positive integer $s$ for which $x^s = e$ is equal to the number of elements in $\langle x \rangle$. I used Lagrange's theorem to calculate this number, which is (by the theorem) the number of elements in $H$ multiplied by the number of cosets of $H$. Exactly $n \times 2$. – brick Nov 10 '14 at 22:37
  • i only ask because in theory it could be a multiple of s – Jack Armstrong Nov 10 '14 at 23:16