Subgroup which is generated by odd elements
Solution 1:
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.
Solution 2:
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.