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I am trying to understand the following:

Let $H$ be a subgroup of index 2 in a group $G$. Show that $g^2 \in H$ for all $g \in G$.

I don't really know how to approach this. I can see that $H$ is normal since there are only 2 cosets, but then I'm stuck. Could somebody give me a hint?

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Since $i_G (H) = 2$, $H \unlhd G$ as you correctly pointed out. The result is a lemma to the following proof.

Consider $h \in H$. Then $h^2 \in H$ by closure in $H$. Now consider some $g \in G$ s.t. $g \notin H$. Since $H \unlhd G$, then $G$ is partitioned by the cosets $He$ and $Hg$ i.e. $G = He \cup Hg$ and $He \cap Hg = \emptyset$.

Now $Hg Hg = Hgg = Hg^2$, once again since $H \unlhd G$, and thus $g^2 \in Hg^2$.

It follows that $Hg^2 \cap Hg = \emptyset$ (since $g \neq e$, otherwise $g \in H$), and there are only two cosets of $H$ in $G$. Thus it follows that $Hg^2 = He$ and hence $g^2 \in H$.

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  • $\begingroup$ Just to clarify: Alot of answers are telling me that anything squared is the identity for groups of order two. But how can I say anything about the order of the elements in $G/N$? Is it because Every group of a prime order is cyclic and isomorphic to $Z_2$? Or is there a simpler explanation I am missing? $\endgroup$ – joakimb Mar 9 '18 at 12:43
  • $\begingroup$ What do you mean by "anything squared", you mean any $g \in G$? Note that in this case you do not have a group of order 2, rather you have that the number cosets of $H$ in $G$ is 2, denoted by $i_G (H) = 2$. Hope this clarifies. $\endgroup$ – Xandru Mifsud Mar 9 '18 at 12:54
  • $\begingroup$ Okay I understood to what you were referring to - my bad. You know that order of $G/H$ is 2 since $i_G(H) = 2$, since the quotient group is the group of all cosets of $H$ in $G$. Hence the quotient group is cyclic and thus coset generates $G/H$. The result follows - this is an alternate proof to what I provided, to which other users have hinted down below. $\endgroup$ – Xandru Mifsud Mar 9 '18 at 12:59
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Hint: Of the two cosets, which one is $g^2H$?

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Hint:

Obviously if $g\in H$ there's nothing to prove.

But what if $g\notin H$? Then the two cosets are $\{H, gH\}$. Can it be the case that $g^2H=gH$?

Once you conclude that's not possible, $g^2H=H$ says exactly that $g^2\in H$.


Using your idea that $H$ is normal:

$G/H$ is a group of order $2$, so $(gH)^2=H$ for every $g\in G$. But that leads to exactly what you are looking for...

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$G/H$ has order two, so anything squared is the identity.

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You understand the notion of normal subgroup, so I'm assuming that you know what a quotient is.

$G/H$ is a group of order $2$, therefore any element has order $1$ or $2$ there. What does it mean in terms of cosets that $$(gH)^2=H\text{ (the identity in the quotient)?}$$

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