# Order of nontrivial elements is 2 implies Abelian group

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

• 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. Commented Jan 11, 2013 at 1:19
• HI Calvin. I do not understand your proof. Are a and b supposed to be arbitrary elements in a group having the property that every (non-identity) element has order 2? I understand the second equation follows from associativity but I do not understand anything else in your proof. Commented Jul 4, 2015 at 23:02
• @GeoffreyCritzer Yes. $a^2 = b^2 = 1$. So $ab = a^2 * ab * b^2$. Commented Jul 14, 2015 at 0:46
• Possible duplicate of Every element of a group has order $2$. Why, intuitively, is it abelian? Commented Sep 17, 2016 at 8:33

Taking inverses reverses the order of multiplication, so if every element is its own inverse multiplication must be commutative.

• That is a nice way of saying it. Commented Jan 11, 2013 at 1:48
• Wow, this is really nice. Commented Jan 11, 2013 at 4:02

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?

• 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 ... Commented Jan 11, 2013 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 $G' \subseteq N$. Commented Jan 13, 2013 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 $H$. Commented Jan 13, 2013 at 7:05

$[a,b]=1$ 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 $H \unlhd G$).

• 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 $r, f$. Commented Jan 11, 2013 at 2:12
• @CalvinLin Try the contrapositive. Commented Jan 11, 2013 at 2:26

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 $ab=ba$.

• 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. Commented Jan 11, 2013 at 10:00

Lemma: Let $$G$$ be a group. Let $$g\in G\setminus \{e\}$$. Then $$|g|=2$$ if and only if $$g=g^{-1}.$$

Proof: Suppose $$|g|=2$$. Then $$g^2=e$$, so

\begin{align} g&=ge\\ &=g(gg^{-1})\\ &=g^2g^{-1}\\ &=eg^{-1}\\ &=g^{-1}. \end{align}

Now suppose $$g=g^{-1}$$. Then $$g^2=gg=gg^{-1}=e$$. But $$g\neq e$$. Hence $$|g|=2.$$ $$\square$$

Lemma: Let $$g,h$$ be in a group $$G$$. Then $$(gh)^{-1}=h^{-1}g^{-1}.$$

Proof: We have that inverses are unique and $$(gh)(h^{-1}g^{-1})=geg^{-1}=gg^{-1}=e.\, \square$$

Theorem: Suppose $$a^2=e$$ for all $$a\in G$$, where $$G$$ is a group. Then $$G$$ is abelian.

Proof: Let $$g,h\in G$$. Then

\begin{align} gh&=(gh)^{-1}\\ &=h^{-1}g^{-1}\\ &=hg. \end{align}

But $$g,h\in G$$ were arbitrary. Hence $$G$$ is abelian. $$\square$$

• There is a typo in the proof of the first lemma, the order of g should be |g|=2. Commented Feb 3 at 17:46
• Thank you, @Lazy_Butter :) Commented Feb 3 at 17:47