The task is as follows: Let $G$ be a finite group.
a) If $|G|$ is divisible by $2$, then there is an element of $G$ with order $2$.
b) If $G$ is abelian group and $|G|$ is divisible by ${2}^{n}$ ($n \in \mathbb{N}$), then there is a subgroup of $G$ with order ${2}^{n}$ .
I have already solved part a). But I need help to prove part b).
The fundamental theorem of abelian groups is not known yet.
It is advised to prove it via induction:
Induction start: $n=0$. Then the trivial subgroup $\{e\}$ is a subgroup of $G$.
Induction step: Let b) be true for an $n \in \mathbb{N}$ and let $G$ be a finite group with $|G|$ divisible by ${2}^{n+1}$. There is a subgroup $U$ with order ${2}^{n}$. Because $G$ is abelian $U$ is a normal subgroup of $G$ and therefore $G/U$ is a group.
Part a) says that there is an element $\overline{b} \in G/U$ with $\operatorname{ord}(b) = 2$.
That means: ${\overline{b}}^{2} = U$ (Operation in $G/U$)
So $\langle${ c : $\overline{b}=\overline{c}$}$\rangle$ (Operation in $G$) is a subgroup of $G$ with |$\langle${ c : $\overline{b}=\overline{c}$}$\rangle$| = $2*{2}^{n}$ = ${2}^{n+1}$
Can someone please review my proof?
Best Answer
The notation $\langle c \in \overline{b} \rangle$ does not mean anything meaningful.
Instead, use a) to find an element $g \in G$ of order $2$. Since $G$ is abelian, $U := \langle g \rangle \subseteq G$ is a normal subgroup. Now apply the induction hypothesis to the group $G/U$, which has order $\mathrm{ord}(G)/2$.
You can use the correspondence theorem for subgroups.