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Defining a group $(G,*)$ where $a^2=e$ with $e$ denoting the identity class....

I am to prove that this group is commutative. To begin doing that, I want to understand what exactly the power of 2 means in this context. Is the function in the group a power or something?

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marked as duplicate by Git Gud, Olivier Bégassat, Magdiragdag, Davide Giraudo, Thomas Feb 5 at 19:16

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For all $a$.... –  Paze Feb 4 at 18:11

5 Answers 5

The trick with these types of problems is to evaluate the 'product' of group elements in two different ways.

So for this problem, we interpret $(ab)^2$ two different ways, where $a,b \in G$.

First, we have this rule in $G$ that an element 'squared' is the identity. So we know that $$ (ab)^2=e $$ But $$ (ab)^2=abab $$ Also note that $$ e=e\cdot e=a^2b^2 $$ So we must have $$ a^2b^2=abab $$ But then that gives us $$ \begin{align} a^{-1}a^2b^{2}b^{-1}&=a^{-1}ababb^{-1}\\ ab&=ba \end{align} $$ since $a,b \in G$ were arbitrary, $G$ is commutative. Later we use the same trick for rings by evaluating $(a+b)^2$ two different ways.

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$$a^2 = a*a$$ where '$*$' is the group operation.

In general, $$a^n = \underbrace{a*a*\cdots * a}_{\large n\;\text{times}}$$

For the proof that $G$ must be commutative, if you're stuck, look at the product of two elements. To simplify, I'll omit the '$*$' symbol for the group operation and simply use juxtaposition of two elements to denote the group operation.

Take $a, b \in G.$ Then we know that $a^2 = b^2 = e.$ Since $G$ is a group, $ab \in G$, since $G$ must be closed under $*$. Furthermore $(ab)^2 = e$, since $a*b \in G$. So we can see that $$(ab)^2 = abab = e$$

Now, left multiply each side of the equation by $a$, and right multiply each side of the equation by $b$: that gives us $$\begin{align} abab & = e \\ \\ \iff a(abab)b &= aeb \\ \\ \iff a^2(ba) b^2 &= ab \\ \\ \iff ebae &= ab \\ \\ \iff ba &= ab.\end{align}$$

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Okay, let me give it a few minutes and then hit you back if I need some help solving the problem :) Thanks –  Paze Feb 4 at 18:13
    
Let me know if you're stuck! ;-) –  amWhy Feb 5 at 17:33

If $a^2 =e$ for all $a\in G$, then $a^{-1}(a^2) = a^{-1}e = a^{-1}$ for all $a\in G$, but since $a^{-1}(a^2) = (a^{-1}a)a =ea = a$, we have $a^{-1} = a$ for all $a \in G$. If we now take any $a, b \in G$, we see that $(ab)^{-1} = ab$. But $(ab)^{-1} = b^{-1}a^{-1} = ba$; thus $ab = ba$ for all $a,b \in G$ and so $G$ is commutative. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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Can you elaborate on the $ (ab)^{-1} = b^{-1} a^{-1} = ba $ part? I'm lost there. –  Paze Feb 4 at 20:47
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@ Paze: observe that $(b^{-1}a^{-1})(ab) = e$, so $(ab)^{-1} = b^{-1}a^{-1}$. Now use the fact we established that $a^{-1} = a$ for all $a$ to see that $b^{-1}a^{-1} = ba$. Does that clarify? If not, please let me know! Regards. –  Robert Lewis Feb 4 at 21:14

Let $x,y \in G$. Then $(xy)^2=e$, and $x^2 y^2=e e =e$, whence $xyxy=xxyy$. By left-cancellation of $x$ and right-cancellation of $y$, we get that $yx=xy$. Hence $G$ is commutative.

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$ab=aeb=a\left(ab\right)\left(ab\right)b=\left(aa\right)ba\left(bb\right)=ebae=ba$

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