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Let $G$ be a group such that $K = \{x^2 \mid x\in G \}$ is a subgroup of $G$.

(a) If $H$ is a subgroup of $G$ with index $2$ show that $K\subset H$.

(b) Show that the number of subgroups in $G$ with index $2$ is equal to the number of subgroups in $G/K$ with index $2$.

Let $k\in K$ then there exist $x\in G$ such that $k=x^2$. (i) If $x\in H$, then $k=x^2\in H$ (ii) If $x$ is not in $H$ then $G=${$H,xH$} and because $xH\neq x^2H$ so $x^2H=H$ and $k=x^2\in H$

Can I write this like that? or is there something wrong for (a) and for (b). I didn't find anything, help me please

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    $\begingroup$ Did you mean for $K$ to be the subgroup generated by those elements? Otherwise the second question does not really make sense. $\endgroup$ – Tobias Kildetoft Apr 4 '17 at 17:29
  • $\begingroup$ @TobiasKildetoft I took the expression for $K$ at face value and indeed for some groups $K$ is a subgroup, among these gorups (GAP notation): $C3, C5, C7, C3 x C3, A4, C15, C7 : C3, SL(2,3), C3 x A4, (C2 x C2 x C2 x C2) : C3, (C2 x C2 x C2) : C7, A5, ((C2 x D8) : C2) : C3, A4 x A4, PSL(3,2), ((C2 x C2 x C2) : C7) : C3, PSL(3,2), GL(2,4), A6, (C2 x C2 x C2) : PSL(3,2), A7, A8$. What these groups have in common is a mystery to me. $\endgroup$ – Marc Bogaerts Apr 5 '17 at 18:15
  • $\begingroup$ @Marc I think you have an error in your code, as this is never a subgroup in a nonabelian simple group, since when it is, it is normal and clearly proper and nontrivial. $\endgroup$ – Tobias Kildetoft Apr 5 '17 at 18:23
  • $\begingroup$ @TobiasKildetoft That's what I tought too, but a closer inpection shows that in all those cases we have $K = G$. I suspect that these groups have no subgroups of index two. $\endgroup$ – Marc Bogaerts Apr 5 '17 at 18:33
  • $\begingroup$ @Marc How can all elements in a group of even order be squares? Take an element whose order is a power of $2$ as large as possible. If this is the square of an element, what would the order of that element be? $\endgroup$ – Tobias Kildetoft Apr 5 '17 at 18:38
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Your proof of point a) is correct albeit that $x^2H \neq xH$ should be explained. Indeed $x^2 = xh \implies x = h \implies x \in H$, contradiction. For the proof of b) we first have to prove that $K$ is normal, the rest is simply a consequence of the "fourth" isomorphism theorem (see point nr. 3). We have $g^{-1}kg = g^{-1}x^2g = g^{-1}xgg^{-1}xg = (g^{-1}xg)^2$.

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  • $\begingroup$ But $K$ isn't a subgroup in general, so (b) doesn't make sense. $\endgroup$ – Derek Holt Apr 6 '17 at 10:25
  • $\begingroup$ I understood that the OP's question was ill formulated and should read Let $G$ be a group such that .... Such groups exist like all abelian groups, some dihedral groups, the quaternion group and many more. $\endgroup$ – Marc Bogaerts Apr 6 '17 at 10:47
  • $\begingroup$ Yes I am sure you are right, but at he moment it simply states that $K$ is a subgroup of $G$, which is not true in general. $\endgroup$ – Derek Holt Apr 6 '17 at 15:10
  • $\begingroup$ @DerekHolt I fixed it, is it okay now? $\endgroup$ – Marc Bogaerts Apr 7 '17 at 10:04
  • $\begingroup$ Yes that's fine now! $\endgroup$ – Derek Holt Apr 7 '17 at 11:04

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