# Let $A$ be a group, where $a^2=1$, a belongs to $A$. Prove that this group is commutative. [duplicate]

Let $A$ be a group, where $a^2=1$ and $a$ belongs to $A$. Prove that this group is commutative.

Thank you for help.

## marked as duplicate by Watson, Arnaud D., Martin Sleziak, Adam Hughes, LeucippusJan 16 '17 at 1:11

• This has to be a duplicate. – Thomas Andrews Apr 3 '14 at 5:41
• It seems a lot of people here are assuming properties of exponents. Obviously for regular numbers, it's true $(xy)^2 = x^2y^2$, but this property isn't one of the fundamental properties of a group. Is there some theorem that says this is always the case for all groups? – Jared Apr 3 '14 at 5:56
• @Jared: $(xy)^2 = x^2y^2$ is not the case for all groups. For $(xy)^2 = x^2y^2$ then $xyxy = x^2y^2$ implying $xy = yx$, so $(xy)^2 = x^2y^2$ if and only if the group is abelian. – Robert Lewis Apr 3 '14 at 6:05
• @RobertLewis I didn't think so (was specifically thinking about invertible matrices). I thought your proof used this as well, but now I see $(xy)^{-1} = y^{-1}x^{-1}$ is certainly correct since $x(yy^{-1})x^{-1} = 1$. – Jared Apr 3 '14 at 6:11
• You need to be more precise. Does $a^2=1$ stand for every $a\in A$ ? – Sylvain Biehler Apr 3 '14 at 9:21

Let $a,b$ be elements of the group. We want to show that $a * b = b * a$. But this is equivalent to $a * b * a = b$ which is equivalent to $(a * b) * (a * b) = 1$, which is true by our hypothesis that $x * x = 1$ for every $x$ in the group.

• I see it now, it would have been easier though (for me) if you had written $(a*b*a)*b = (b)*b$. – Jared Apr 3 '14 at 5:49

We have: $x^2 = e$, and $y^2 = e$. So: $(xy)^2 = e = e*e = x^2y^2$. So $(xy)^2 = x^2y^2$. Thus: $xyxy = xxyy$. So: $x^{-1}xyxyy^{-1} = x^{-1}xxyyy^{-1}$. Since $x^{-1}x = yy^{-1} = e$, and $ex = x$, $ey = y$, you have : $yx = xy$ for all $x, y$ proving the group commutative.

• How can you claim $(xy)^2 = (x*y)*(x*y) = x*x*y*y$ without assuming commutativity? – Jared Apr 3 '14 at 5:52
• OK, I think I understand now. $(xy)*(xy) = 1$, which must also equal $(x*x)*(y*y) = 1*1 = 1$. But it seems like it's easier to simply state $x*y*x*y = x*(y*x)*y$ and if that equals $x*x*y*y$ (which it does since they are both the identity) then it must be valid to swap the middle term, i.e. $x * (y * x)* y = x * (x*y)*y$ therefore $y*x = x*y$. – Jared Apr 3 '14 at 6:06

Here's my take on it:

Since $a^2 = 1$ for all $a \in G$, $a = a^{-1}$ for all $a \in G$. Let $x, y \in G$; then $xy = (xy)^{-1} = y^{-1}x^{-1} = yx$, since $x = x^{-1}$, $y = y^{-1}$, and $xy = (xy)^{-1}$. $G$ is indeed abelian. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

It's not true. Choose for example: $$a_1=\pmatrix{0 & 1\\1&0} \text{ and } a_2=\pmatrix{1 &0\\0&-1}.$$ For both you have $a_k^2=1$, but the generated group is not commutative.

• Did he say anything about generators? $(a_1a_2)^2\neq 1$ so your group doesn't qualify. – Thomas Andrews Apr 3 '14 at 5:36
• @ThomasAndrews no, I just thought I construct a group as a counterexample. It just happened that my example also generates a group. Is my reasoning wrong?...oh I see – draks ... Apr 3 '14 at 5:38
• Well, the question is a mess, but almost every algebra book in existence asks a question: If every element of a group, when squared, is the identity, then that group is commutative. That's how every other person here read the question. – Thomas Andrews Apr 3 '14 at 5:40
• @ThomasAndrews looks like... – draks ... Apr 3 '14 at 5:41

I think this is pretty straigtforward. All you need to use is associativity. We wish to prove $a*b = b*a$. First multiply both sides by $a$:

$$a*(a*b) = a*(b*a) \\ (a*a)*b = (a*b)*a \\ b = a*b*a$$

Now, multiply both sides by $b$: $$b * (a*b) = b*(b*a) \\ b*a*b = (b*b)*a \\ b*a*b = a$$

Now you can plug in $a = b*a*b$ into $b = a*b*a$:

$$b = (b*a*b) * b * a \\ b = b*a*(b*b)*a\\ b = b*(a*a) \\ b=b \text{, q.e.d.}$$

edit as per the comment, here is my edited proof:

I can prove it through the following: \begin{align} b =& b*(a*a) \\ b =& b *a* (b*b) *a \text{, inserted } 1 \text{, between the } a\text{'s} \\ b =& (b*a*b)*b*a \\ b*(b*a) =& (b*a*b)*b*a * b*a\\ (b*b)*a =& (b*a*b)*\left((b*a)*(b*a)\right)\text{, since } (b*a)*(b*a) = 1 \\ a =& b*a*b \\ a*b =& (b*a*b)*b = b*a \end{align}

• This isn't a proof. You've started out assuming the thing you wanted to show, and then deduced something that's also true from that (possibly false assumption). One can 'prove' many things by assuming something that needn't be true, so in a mathematical proof you should only start with what you know is true (i.e., not the statement you wish to prove), and then deduce the thing you want to prove. Your argument would probably work backwards however, since $b=b$ is certainly true... – ah11950 Apr 3 '14 at 7:14
• I'm not totally sure you're right, but I definitely see your logic. I guess I proved $\left(a*b = b*a \wedge a*a=1\wedge b*b=1 \right)\rightarrow b = b$ (and I can certainly see how that's a trivial thing, since $\text{anything} \rightarrow true$ is a tautology). – Jared Apr 3 '14 at 7:24
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