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If the order of all nontrivial elements in a group is 2, then the group is Abelian. I know of a proof that is just from calculations (see below). I'm wondering if there is any theory or motivation behind this fact. Perhaps to do with commutators?


Proof: $a \cdot b = (b \cdot b) \cdot (a \cdot b) \cdot (a \cdot a) = b \cdot (b \cdot a) \cdot (b\cdot a) \cdot a = b \cdot a$.

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It's kind of an odd thing isn't it. Maybe it could help you to understand how counterexamples exist for groups of exponent p > 2 and how these counterexamples fail when p becomes 2. –  Myself Jan 11 '13 at 1:19
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4 Answers

up vote 7 down vote accepted

As every non-identity element has order two, $a^{-1} = a$ for any element of the group. Therefore $$[a, b] = aba^{-1}b^{-1} = abab = (ab)^2 = e.$$ Hence the group is abelian. Is this too calculationy?

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Haha thanks! I knew I was missing something as obvious as that. I thought the commutator subgroup was tricky, but completely forget that I could just do that ... –  Calvin Lin Jan 11 '13 at 1:19
    
As a learner, the following idea helps me: commutator subgroup is exactly what you want to force to be identity in order to make a group abelian. For example, we see that if $N \unlhd G$. Then $G/N$ is abelian if and only if $G' \subseteq N$. –  GYC Jan 13 '13 at 7:02
    
Plus, I agree that learning/using more vocabulary (e.g. "commutator subgroups") is difficult. However, when you deal with more difficult problems, extra vocabulary makes those tedious steps "obvious." A typical example is as follows: let $H \leqslant G$. We see that $N_{G}(H) = \{g \in G : H^{g} = H\}$ is a subgroup, just by looking at it as a stabilizer subgroup of conjugation action of $G$ on $H$. –  GYC Jan 13 '13 at 7:05
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$[a,b]=1$ for all $a,b\in G$ if and only if $G$ is abelian. You proved that $[a,b]=a^{-1}b^{-1}ab=1$ above - this is the connection to commutators.

I don't know of any strong motivation behind this fact aside from, I guess, knowing that any nonabelian group must have an element of order $>2$. I think that it is just a standard exercise.

It may interest you motivationally to prove that $G/H$ is abelian if and only if $G'\leqslant H$ (if $H \unlhd G$).

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Any hints on how to show that any non-abelian group has an element of order >2? If $ab\neq ba$, we could still have $abab=e$. Am thinking of $S_3$ with $r, f$. –  Calvin Lin Jan 11 '13 at 2:12
    
@CalvinLin Try the contrapositive. –  Alexander Gruber Jan 11 '13 at 2:26
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The idea of this approach is to work with a class of very small, finite, subgroups $H$ of $G$ in which we can prove commutativity. The reason for this is to be able to use the results like Cauchy's theorem and Lagrange's theorem.

Consider the subgroup $H$ generated by two distinct, nonidentity elements $a,b$ in the given group. The group $H$ consists of strings of instances of $a$ and $b$. By induction on the length of a string, one can show that any string of length 4 or longer is equal to a string of length 3 or shorter.

Using this fact we can list the seven possible elements of $H$: $$1,a,b,ab,ba,aba,bab.$$ By (the contrapositive of) Cauchy's Theorem, the only prime divisor of $|H|$ is 2. This implies the order of $H$ is either $1$, $2$, or $4$.

If $|H|=1$ or $2$, then either $a$ or $b$ is the identity, a contradiction.
Hence $|H|$ has four elements. The subgroup generated by $a$ has order 2; its index in $H$ is 2, so it is a normal subgroup. Thus, the left coset $\{b,ba\}$ is the same as the right coset$\{b,ab\}$, and as a result $ab=ba$.

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After establishing $|H|=4$ we could have also said $H$ is a p-group so it has nontrivial center, and $H/Z(H)$ is cyclic, so $H$ is abelian. –  peoplepower Jan 11 '13 at 10:00
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Taking inverses reverses the order of multiplication, so if every element is its own inverse multiplication must be commutative.

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That is a nice way of saying it. –  Alexander Gruber Jan 11 '13 at 1:48
    
Wow, this is really nice. –  Michael Albanese Jan 11 '13 at 4:02
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