Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.

Question

Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.

This question is from group theory in Abstract Algebra and no matter how many times my lecturer teaches it for some reason I can't seem to crack it.

(Please note that $e$ in the question is the group's identity.)

Here's my attempt though...

First I understand Abelian means that if $g_1$ and $g_2$ are elements of a group $G$ then they are Abelian if $g_1g_2=g_2g_1$...

So, I begin by trying to play around with the elements of the group based on their definition...

$$(g_2g_1)^r=e$$ $$(g_2g_1g_2g_2^{-1})^r=e$$ $$(g_2g_1g_2g_2^{-1}g_2g_1g_2g_2^{-1}...g_2g_1g_2g_2^{-1})=e$$

I assume that the $g_2^{-1}$'s and the $g_2$'s cancel out so that we end up with something like,

$$g_2(g_1g_2)^rg_2^{-1}=e$$ $$g_2^{-1}g_2(g_1g_2)^r=g_2^{-1}g_2$$

Then ultimately...

$$g_1g_2=e$$

I figure this is the answer. But I'm not totally sure. I always feel like I do too much in the pursuit of an answer when there's a simpler way.

Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra.

Answer 1

Hint: Take $(ab)^2=1$ and multiply both sides on the right with $b$, then again on the right with $a$.

Answer 2

For any $g, h \in G$, consider the element $g\cdot h\cdot h\cdot g.~$ Since $g^2 = g\cdot g= e$ for all $g \in G$, we find that $$g\cdot h\cdot h\cdot g = g\cdot(h\cdot h)\cdot g = g\cdot e\cdot g = g\cdot g = e.$$ But, $g\cdot h$ has unique inverse element $g\cdot h$, while we have just proved that $(g\cdot h)\cdot (h\cdot g) = e$, and so it must be that $g\cdot h = h\cdot g$ for all $g, h \in G$, that is, $G$ is an abelian group.

Answer 3

Whenever you have a condition $g^2=e$ in a group, it's equivalent to $g=g^{-1}$ (multiply both sides by $g^{-1}$).

In this case, it applies to every element of the group, so you can add or remove inverses from any expression freely. So the proof is simply $$ab=(ab)^{-1}=b^{-1}a^{-1}=ba.$$

Answer 4

Then for all $a,b \in G$: $$ab = (bb)ab(aa) = b(baba)a = ba.$$

Answer 5

Hint: Note that $g_1g_2=g_2g_1$ if and only if $g_1g_2g_1^{-1}g_2^{-1}=e$ (Why?), and that $g^{-1}=g$ for all $g\in G$ (Why?).

Answer 6

Alternatively, the map $$\begin{align*}f:G&\rightarrow G\\x&\mapsto x^{-1}(=x)\end{align*}$$ is an automorphism of $G$ and so $G$ is Abelian!

Answer 7

Proof: let for all $a,b$ in group $G$. claim that To show $ab=ba$ a commutative. By using a fact that $a\cdot a=b\cdot b=(ab)\cdot(ab)=e$. since $(ab)^2=a^2\cdot b^2=e\cdot e=e$. We have $ab\cdot ab=e$. Multiplying on the right by $ba$, we obtain \begin{align} ab\cdot ab\cdot ba &= e\cdot ba\\ ab\cdot a(b\cdot b)\cdot a &= ba\\ ab\cdot a\cdot b^2\cdot a &=\\ ab\cdot a\cdot e\cdot a &=\\ ab\cdot a\cdot a &=\\ ab\cdot e &=\\ ab &= ba, \end{align} for all $a,b$ in $G$. since $G$ is abelian group. This is proved last.

Answer 8

Another proof is by contradiction.

Let G be a group with operation *. You want to show that: $(\forall g \in G:g^2=e)\implies(G\text{ Abelian}\Leftrightarrow \forall x,y \in G: x*y = y*x)$.

(where $g^2$ is shorthand for $g*g$)

Suppose by contradiction that the group is not Abelian, i.e. that ($\exists x,y\in G: x*y\neq y*x)$. Now multiply on the left by $x$ and on the right by $y$.

You get $x^2*y^2 \ne (xy)^2$. But then it means that $e*e \neq e$ which is a contradiction.

Answer 9

By construction:

$\quad\begin{align*} (ab)(ab) &= e = a(bb)a \\ \require{cancel}\cancel{(ab)}(ab) &= \cancel{(ab)}(ba) \qquad\text{by associativity followed by cancellation}\\ ab &= ba \end{align*}$

Hence, the group is Abelian.

Answer 10

$(ab)^{2}=e $($\because$ $a,b \in G, ab\in G$, due to the closure property of group axiom )$\implies (ab)(ab)=e\tag 1$ Pre multiply $a$ on both sides also post multiply $b$ on both sides. The equation (1) becomes, $aababb=ab\tag2$($\because$ by associativity and self invertible property. $ba=ab\tag3$

Answer 11

Here's an approach using character theory.

Let $\chi\in\text{Irr}(G)$. It suffices to show that $\chi(1)=1$. We have $$\chi(1)=\frac{1}{|G|}\sum_{g\in G}\chi(1)=\frac{1}{|G|}\sum_{g\in G}\chi(g^2).$$ Hence, $\chi(1)$ coincides with the Frobenius-Schur indicator of $\chi$, which is necessarily equal to $1$, $0$ or $-1$. Since $\chi(1)\geq1$ then $\chi(1)=1$ and we are done.

Answer 12

$$xyx^{-1}y^{-1}=(xyx^{-1})^2x^2(x^{-1}y^{-1})^2$$

Answer 13

We have for all $a,b\in G$,

$$\begin{align} \color{red}{abab}&=(ab)^2\\ &=e\\ &=ee\\ &=a^2b^2\\ &=\color{red}{aabb}, \end{align}$$

so that

$$\begin{align} ba&=e(ba)e\\ &=(a^{-1}a)ba(bb^{-1})\\ &=a^{-1}(\color{red}{abab})b^{-1}\\ &=a^{-1}(\color{red}{aabb})b^{-1}\\ &=(a^{-1}a)ab(bb^{-1})\\ &=e(ab)e\\ &=ab. \end{align}$$

Answer 14

Let $a,b\in G$: \begin{align} ab\cdot (ab)^{-1} &= ab\cdot b^{-1}a^{-1}\\ &= ab\cdot ba\quad\quad \text{(since each element in $G$ is self inverse)}\\ &= a(b^{2})a\\ &= a\cdot e\cdot a\\ &= a^{2}\\ &=e \end{align} This shows $(ab)^{-1}=ba$. Now $(ab)^2=e$, so $ab\cdot ab=ab\cdot ba=e$, which by the cancellation law gives $ab=ba$ for all $a,b\in G$, since $a$ and $b$ were arbitrary. Hence $G$ is abelian as required.