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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 to mean 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...


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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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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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) – BU982T May 20 '13 at 21:52

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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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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HELLO MATHEMATICS student I show for you as a hint to prove the question above and show again clearly to understand.

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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