# Elements of odd order generate a proper subgroup of a group

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!

• Is $G$ abelian? For a general group the odd order elements do not form a subgroup! – Espen Nielsen Sep 28 '15 at 21:22
• Do you mean that the elements of odd order generate a proper subgroup? – Mark Bennet Sep 28 '15 at 21:48
• 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. – Sammy Black Sep 28 '15 at 21:48
• @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. – Espen Nielsen Sep 28 '15 at 21:55
• 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 $H$. – Derek Holt Sep 29 '15 at 7:40

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 $$P\neq G$$.

• 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. – Tobias Kildetoft Feb 23 at 7:15
• You are right, I will edit my answer accordingly. – B.Swan Feb 23 at 16:53