Subgroup which is generated by odd elements Let $G$ be a finite Group, let S=$\{s_1,s_2,s_3 ... |s_i\in G $ has odd order $\}$.
Let $H=\langle S\rangle$
The question is:


*

*Prove that $H$ is normal subgroup of $G$.

*prove that $G/H$ has order $2^k$ for $k \in \mathbb{N}$.
I tried look at $$gHg^{-1}$$ but can't understand why $gHg^{-1} \subseteq H$ probably because I cant understand what unique in H.
Thanks much for help.
 A: Hint:


*

*It is enough to show that, if $h\in G$ has odd order and $g\in G$, then $ghg^{-1}$ has odd order. Note here that conjugation is an isomorphism, hence it preserves the order of elements.

*It suffices to show that each non-identity element of $G/H$ has even order.
A: Others already proved that $H \unlhd G$. But I did not see a proper proof of (2). Here it is: let $\bar{x} \in G/H$ be a non-trivial element, that is $\bar{x} \neq \bar{1}$, hence $x \notin H$, and this impies that $x$ has even order, say $|x|=2^i\cdot k$, with $k$ odd and $i \geq 1$. Then $x^{2^i}$ has odd order (namely $k$), hence $\bar{x}^{2^i} = \bar{1}$ in $G/H$. We conclude that every element of $G/H$ is a $2$-element, hence $|G:H|$ is a power of $2$.
There is a generalization to general sets of primes $\pi$, where the above is the special case $\pi=\{2\}$, see here.
A: *

*Every element of $H$ is generated by a finite product of $s_1, \dots, s_i, \dots$ (note that inverses are already included in the set $S$), so let $h \in H$, then $h=s_{i_1}.. s_{i_n}$ for some $n \in \mathbb{N}$. Let $g \in G$. Then $$g^{-1}hg =  g^{-1}s_{i_1}.. s_{i_n}g = g^{-1}s_{i_1}gg^{-1}s_{i_2}g.. g^{-1}s_{i_n}g=(g^{-1}s_{i_1}g)(g^{-1}s_{i_2}g).. (g^{-1}s_{i_n}g)=s_{j_1}.. s_{j_n}$$ for some indices $j_i$, since $s_i$ and $g^{-1}s_i g$ have the same order (in particular, odd) so there exists a $j$ such that $g^{-1}s_i g = s_j$.

*Suppose that $p$ is an odd prime that divides $|G/H| = [G:H]$. Then there exists an element $x \in G/H$ of order $p$ (Cauchy's theorem). In particular $x \neq 1$ in $G/H$ so if we consider its lateral $xH$ we get an element $\bar{x} \in G$ with odd order not contained in $H$, which is a contradiction. Then there are no odd primes that divide $[G:H]$, which means that $[G:H]=2^k$ for some $k \in \mathbb{N}$.
