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I am stuck with the following question:

Given is a group $G$ with a subgroup $H$ of index $2$, so $\left [ G:H \right ]=2$. I have to show that the elements of odd order of $G$ generate a proper subgroup.

What i know is that $H$ as a subgroup of index $2$ has only $2$ left cosets (and also right too). So i know that this subgroup is normal in $G$. What can i do with the order? Can anybody help me with this exercise, please? Thank you in advance!

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    $\begingroup$ Is $G$ abelian? For a general group the odd order elements do not form a subgroup! $\endgroup$ – Espen Nielsen Sep 28 '15 at 21:22
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    $\begingroup$ Do you mean that the elements of odd order generate a proper subgroup? $\endgroup$ – Mark Bennet Sep 28 '15 at 21:48
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    $\begingroup$ Explicitly, in $S_4$, the symmetric group on $\{1, 2, 3, 4\}$, the elements (in cycle notation) $(1\,2\,3)$ and $(2\,3\,4)$ each have order $3$, but their product (composing from left to right) is $(1\,3)(2\,4)$, which has order $2$. The group $S_4$ does have an index $2$ subgroup, namely $A_4$, but $S_4$ is not abelian. $\endgroup$ – Sammy Black Sep 28 '15 at 21:48
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    $\begingroup$ @MarkBennet That's true, but since we may always add a factor $\mathbb{Z}/2\mathbb{Z}$, this is not a difficult requirement to meet. $\endgroup$ – Espen Nielsen Sep 28 '15 at 21:55
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    $\begingroup$ As has been pointed out, the result is not true, so you there must either be a mistake in the question, or you have not copied it correctly. It is true that the elements of odd order generate a proper subgroup, because all usch elements are contained in $H$. $\endgroup$ – Derek Holt Sep 29 '15 at 7:40
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The elements of odd order generate a subgroup, let's call it $P$, because $1 \in G$ has order $1$, which is odd. So $1 \in P$.

Thus $P$ contains some elements and since we are generating a group, we automatically generate the inverses and all products, so $P$ will be a subgroup.


We still need to show that $P\neq G$ for $P$ to be a proper subgroup.

We know that there is an $H \leq G$ with $[G:H]=2$, so $H\neq G$.

We will show that $P$ lies in $H$.

Theorem 2. If $G$ is a finite group and $N \triangleleft G$ then any element of $G$ with order relatively prime to $[G:N]$ lies in $N$. In particular, if $N$ has index $2$ then all elements of $ G$ with odd order lie in $N$.

Proof: Let $g$ be an element of $G$ with order $m$, which is relatively prime to $[G:N]$. The equation $g^m=e$ gives $(gN)^m=N \in G/N$. Also $(gN)^{[G:N]}=N$, as $[G:N]$ is the order of $G/N$.

So the order of $gN \in G/N$ divides $m$ and $[G:N]$.

These numbers are relatively prime, so $gN=N$, which means $g \in N$.

Now we know that all odd elements lie in $H$. Since $H$ is a subgroup, it is closed, so the whole subgroup generated by odd elements must lie in $H$. Thus $P\neq G$.

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  • $\begingroup$ Most of this could probably be left out since the statement that the elements generate a subgroup is vacuously true since this holds for any subset. On the other hand, including a short proof of the main result cited would make the answer self-contained which is generally preferable. $\endgroup$ – Tobias Kildetoft Feb 23 at 7:15
  • $\begingroup$ You are right, I will edit my answer accordingly. $\endgroup$ – B.Swan Feb 23 at 16:53

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