2
$\begingroup$

Suppose that $G$ is a group with $g^2 = I$ for all $g \in G$. Show that $G$ is necessarily abelian. Prove that if $G$ is finite, then $|G| = 2^k$ for some $k>=0$ and $G$ needs at least $k$ generators.

Well for the first part - G has order two so it must be the cyclic group with two elements. So we have $gI = Ig$ as the only ways of multiplying the element so G has to be abelian?

I don't know how you would go about proving the second part. I can't see how $G$ being finite means $G$ will necessarily have order of $2^k$ and will need at least $k$ generators?

  • 2
    I'm assuming you mean $g^2=I$ for all $g\in G$. Note that the Klein four group is a group with four elements, and every non-identity element has order two, so your conclusion that $G$ has order two is incorrect.2012-10-25
  • 1
    $G$ isn't necessarily of order 2. Each element has order 2, but you could have an group $G=\{a_{1},a_{2},...,g_{n}\}$, where each $a_{i}$ satisfies $a_{i}^{2}=e$. From there, see if you can show $G$ is abelian without making assumptions about its size.2012-10-25
  • 0
    For the second part, you could use the structure theorem for finite Abelian groups.2012-10-25
  • 0
    For the second part, let $\{g_1,g_2,\ldots,g_k\}$ be a minimal generating set of $G$ and use induction on $k$. The case $k=1$ is easy. Let $H = \langle g_1,\ldots,g_{k-1} \rangle$. By induction, $|H| = 2^{k-1}$. Now prove that $G = H \times \langle g_k \rangle$ (or it would be enough just to prove $|G:H|=2$).2012-10-25

1 Answers 1