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?

  • 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

2

For the first part:

You know that for any element $x \in G$ that $x^2 = e$.

There are many ways of turning this into a proof that $G$ is abelian.

Pick any $a, b \in G$. Then $(ba)^2 = e = e*e = b^2 * a^2$. But this says $baba = bbaa$, and we can now cancel $a$'s from the right and $b$'s from the left to conclude $ab = ba$.

Alternatively $x^2 = e$ iff $x = x'$. That is, if and only if every element is equal to its own inverse.

Pick any $a, b \in G$. Then $ab = (ab)' = b'a' = ba$.

As a third approach, note that $(ab)^2 = e$ means $abab = e$, after which we multiply both sides by $ba$ on the right and reach the desired conclusion shortly thereafter.

For the second part:

Suppose a prime $p > 2$ divided the order of $G$. Are you familiar with the Sylow theorems?

  • 0
    Frankly, I'm surprised to see these two problems appear together. The first can be assigned within the first week of a course on Group Theory, while the second (if we are to use Cauchy's Theorem or its generalization to Sylow's first theorem) might not appear until much later in the course; at the least product groups, (normal) subgroups, and Lagrange's Theorem would probably have been covered.2012-10-25