Let $A$ be a group, where $a^2=1$, a belongs to $A$. Prove that this group is commutative. Let $A$ be a group, where $a^2=1$ and $a$ belongs to $A$. Prove that this group is commutative.
Thank you for help.
 A: 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.
A: 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.
A: 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!!!
A: 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.
A: 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}
