I was working on a finite group $G$ in which every element different from the identity element has an order of $2$, and my goal is to prove that $|G|=2^r,r\in\mathbb{N}$. I did find a proof and I want to know whether it is true or not.
Used lemma:
- for every subgroup $H\neq G$,$\forall a\in G\backslash H,K=H\cup aH$ is a subgroup of $G$ of order $2|H|$
My attempt: Let $n$ be the order of $G$, Let $(a_k)_{k\in\{{0,..,n-1}\}}$ the elements of $G$, with $a_0$ the identity element. Let $(H_k)_{k\in\{{0,..,n-1}\}}$ defined in the following way:
- $H_0=\{a_0\}$
- for $k\in\mathbb\{0,..,n-2\}$: $H_{k+1}=H_k\cup a_{k+1}H_k$.
I will prove by induction that $\forall k\in\{0,..,n-1\},\forall p\in\{0,..,k\},a_p\in H_k$
- Base Case $(k=0):\ $we have $a_0\in H_0\implies\forall p\in\{0,..,k\},a_p\in H_k$
- Let $k\in \{0,..,n-2\}/\forall p\in\{0,..,k\},a_p\in H_k$, we have $$ H_k\subset H_{k+1} \text{ and }a_{k+1}\in H_k \text{ from the definition of } (H_k) \\ \implies \forall p\in\{0,..,k+1\},a_p\in H_{k+1} $$ Hence our result is proved, we can deduce that $|H_{n-1}|\ge n\implies |H_{n-1}|\ge|G|$
Now I will prove (by induction again) that : $\forall k\in\{0,..,n-1\} \ H_k$ is a subgroup of $G$ and $\exists r \in \mathbb{N}/|H_k|=2^r$
- Base Case: $H_0$ is a (trivial) subgroup of $G$ and $|H_0|=2^0$
- Let $k\in \{0,..,n-2\}/ \ H_k$ is a subgroup of $G$ and $\exists r \in \mathbb{N}/|H_k|=2^r: \ $
if $a_{k+1} \in H_k$, then: $$ a_{k+1}H_k=H_k \text{ since } H_k \text{ is a group} \implies H_{k+1}=H_k$$ Otherwise, by our lemma, we have $H_{k+1}$ is a subgroup of $G$ of order $2|H_k|=2^{r+1}$.
In both cases, $H_{k+1}$ is a subgroup of $G$ and $\exists r \in \mathbb{N}/|H_{k+1}|=2^r$.
So this result is also proved, and we can deduce that $H_{n-1}$ is a subgroup of $G$ and $|H_{n-1}|=2^r,r\in \mathbb{N}$, So $H_{n-1}\subset G$, but $|H_{n-1}|\ge |G|\implies H_{n-1}=G$
Conclusion: $$|G|=n=2^r,r\in\mathbb{N}$$
So that was my proof, and my main concern here is the validity of this form of induction (and my proof in general), and thanks for your help.
By Cauchy's theorem if the order of the group were divisible by an odd prime $p$ it would have an element of order $p$.