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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?

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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. – Bey Oct 25 '12 at 1:31
$G$ 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. – user123123 Oct 25 '12 at 1:34
For the second part, you could use the structure theorem for finite Abelian groups. – Bey Oct 25 '12 at 1:40
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 $|G:H|=2$). – Derek Holt Oct 25 '12 at 8:25

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?

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Perhaps I'm not seeing something about your hint to OP for the second part. Obviously |G| is even since the order an element divides the order of the group. But the issue in applying Sylow theorems is that we don't yet know anything about the multiplicity of 2 in |G|. – Bey Oct 25 '12 at 1:45
If a prime $p$ divides the order of $G$, then there is an element in $G$ of order $p$; call it $g$. If $p > 2$, then in particular $g^2 \neq e$. – Benjamin Dickman Oct 25 '12 at 2:10
Then @B., a perhaps more basic hint would be Cauchy's Theorem and not Sylow ones. – DonAntonio Oct 25 '12 at 2:41
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. – Benjamin Dickman Oct 25 '12 at 5:52

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