Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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. –  Bey Oct 25 '12 at 1:31
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. –  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
add comment

1 Answer

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?

share|improve this answer
1  
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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.