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Here is more details about the question:

Let $G$ be a finite group and suppose that all (non-identity) elements have order $2$.

(a) Prove that $G$ is abelian.

(b) Let $a, b$ be two distinct non-identity elements in $G$. Show that the subgroup generated by $a$ and $b$ consists of $4$ distinct elements $e, a, b, ab$ and hence |$\langle a, b\rangle$| $= 4$.

(c) Prove by induction on n that for any set of elements $a_1, a_2, ...a_n$ the subgroup generated by them is of order a power of 2. Deduce that |G| is a power of $2$.

I have finished (a),(b):(a) is simple, and in (b) I define the subgroup $S=\{a^ib^j,i,j\in \{0,1\}\}$ then reach the result but I not very confident. in (c) ,it is about Lagrange's theorem and I think we should use $o(g)$ divides $|G|$, but I am not sure precisely how to reach it.

Stéphane Jaouen
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iwueh
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1 Answers1

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Your idea of introducing $S$ into (b) is a good idea. You just have to prove that $S=\{e,a,b,ab\}$ and $|S|=4$. For example, $ab\neq a$ otherwise $b=e$.

It's a good idea for induction in (c). Let's deal with the case $S=\langle a,b,c \rangle$.

Case 1) $c=ab$ hence $|S|\in\{1,2,2^2\}$;

Case 2) $a\neq e, b\neq e, a\neq b, c\notin \langle a,b \rangle$. Then $g\in \langle a,b,c \rangle \iff g\in\{e,a,b,ab\} \cup\{c,ac,bc,abc\}$. And $|\langle a,b,c \rangle |=2^2\times2=2^3$

And so on...

Stéphane Jaouen
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  • in induction process why we start with the case $S=\langle a,b,c\rangle$ instead of when n=1? – iwueh Feb 18 '24 at 17:13
  • I didn't write to "start the induction with the case $S=<a,b,c>$". Rather than dealing with the general case $H_n\implies H_{n+1}$, I have limited myself to showing that $H_2\implies H_3$ which is general enough to be easily generalized. What is important then is to understand the importance of considering $c \notin <a,b>$ – Stéphane Jaouen Feb 18 '24 at 17:22
  • I see. I still wonder how to complete the proof after proving $H_2 ⟹ H_3$.Thank you so much! – iwueh Feb 18 '24 at 17:35
  • Use my comment 6 hours ago – Stéphane Jaouen Feb 18 '24 at 17:38