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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.

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  • $\begingroup$ What does r stand for? $\endgroup$ – john Oct 5 '17 at 21:03
  • $\begingroup$ I don't think we can assume that $ g_2 g_1 $ has finite order , with $(g_2g_1)^r=e $ $\endgroup$ – john Oct 5 '17 at 21:05

13 Answers 13

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

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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.

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  • $\begingroup$ This is the elegant solution to this problem. I thought I'd post this exact solution on here if everyone did it the long way, but I'm half a year late. (+1) $\endgroup$ – user70962 May 20 '13 at 21:52
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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 $$ab=(ab)^{-1}=b^{-1}a^{-1}=ba.$$

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  • $\begingroup$ What happens after the 2nd equality? $\endgroup$ – Dole Sep 13 '17 at 2:09
  • $\begingroup$ @Dole, 1st equality: addition of an inverse, 2nd equality: formula for inverse of a product, 3rd equality: removal of inverses. Remember in this group, we can add or remove $^{-1}$ from anything, because every element is its own inverse. Does that answer your question? $\endgroup$ – asmeurer Sep 14 '17 at 23:27
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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?).

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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.

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  • 2
    $\begingroup$ I think you meant "therefore" instead of "since" in the last sentence. That $G$ is abelian was to be proved, not given (and BTW, nowhere you've used that it is specifically a group; the argumentation works just as well for monoids that are not groups). $\endgroup$ – celtschk Dec 14 '18 at 8:51
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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.

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    $\begingroup$ Hi, I know this is old, but how can you multiply the left side by $x$ and right side by $y$ and not do the same for both sides? In other words, if I multiply one side by $x$, don't I have to do the same for the other side? I am not sure if I am reading that correctly. $\endgroup$ – Ryan Feb 10 at 1:22
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    $\begingroup$ $\color{red}{x} *x*y*\color{red}{y} \neq \color{red}{x} * y * x * \color{red}{y} $, so to be precise I should have said multiple both sides on the left by $x$ and on the right by $y$. Then applying associativity ($*$ is associative by definition of group) you get $(\color{red}{x} *x)*(y*\color{red}{y}) \neq (\color{red}{x} * y) * (x * \color{red}{y})$ and there you go. I hope it helps. $\endgroup$ – Marco Bellocchi Feb 10 at 9:49
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    $\begingroup$ Thank you for responding to an old post and for the clarification! $\endgroup$ – Ryan Feb 10 at 18:43
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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!

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  • $\begingroup$ This is actually quite neat! There is no circular logic either, which I was initially worried about. (Also, I've edited the post to make the map clearer. Feel free to undo my edit if you wish.) $\endgroup$ – user1729 Dec 14 '18 at 15:35
  • $\begingroup$ @user1729: .This time it looks good. Thanks for the edit! $\endgroup$ – Chinnapparaj R Dec 15 '18 at 1:39
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    $\begingroup$ best answer by far. $\endgroup$ – Ivan Di Liberti Jan 13 at 23:53
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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.

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  • $\begingroup$ This is very direct and nice. Is this proof plausible? $\endgroup$ – Ryan Feb 10 at 1:26
  • $\begingroup$ What makes you think this proof may not be plausible? $\endgroup$ – hchar Feb 11 at 4:24
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Then for all $a,b \in G$: $$ab = (bb)ab(aa) = b(baba)a = ba.$$

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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.

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$(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. $ba=ab\tag3$

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given $g^2=e$ for all $g\in G$ So $g=g^{-1}$ for all $g\in G$ Let,$a,b\in G$ Now $ab=a^{-1}b^{-1} =(ba)^{-1} =ba$ So $ab=ba$ for all $a,b\in G$ .Hence $G$ is Abelian Group.

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

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