Group Theory

This week, we present a guest post by Chu-Wee Lim on Group Theory.

How would you prove the following?

GG is a group such that for every element gG g \in G , we have g2=e g^2 = e . Prove that GG is abelian.

Note by Calvin Lin
6 years, 2 months ago

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12 votes

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Let a,bGa,b \in G. Note that (ab)2=e(ab)^2 = e because abGab \in G. So e=(ab)2=ababe = (ab)^2 = abab Hence aeb=a(abab)baeb = a(abab)b ab=(aa)(ba)(bb)ab = (aa)(ba)(bb) ab=a2bab2ab = a^2bab^2 ab=ebaeab = ebae ab=baab = ba for all a,bGa,b \in G, that is to say GG is abelian.

Jan J. - 6 years, 2 months ago

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We can see it more easily as follows.

The given condition implies g=g1,gGg=g^{-1}, \forall g \in G . Hence

ab=(ab)1=b1a1=baab= (ab)^{-1} = b^{-1}a^{-1}=ba \hspace{5pt} \blacksquare

Abhishek Sinha - 6 years, 2 months ago

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Yes this is a good exercise but here is another one: let GG be a finite abelian group GG is cyclic G=ord(G)\Longleftrightarrow |G| = ord(G). ord(G)= ord(G) = the least positive integer nn such that gGgn=e g \in G \Longrightarrow g^{n} = e.

Samuel Queen - 6 years, 2 months ago

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It says that

gm+n=gmgn=(gm)n, g^{m+n} = g^mg^n = (g^m)^n,

which isn't true. Apart from that, great post. :)

Tim Vermeulen - 6 years, 2 months ago

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Tim, can you clarify what you mean? I do not see that equation written anywhere.

The only relevant statement that I see is the third box, which states that:

For any gGg \in G and m,nZm, n \in \mathbb{Z} , then gm+n=gmgn g^{m+n} = g^m g^n and (gm)n=gmn \left( g^m \right)^n =g^{mn} .

This is a correct statement, and different from what you quoted.

Calvin Lin Staff - 6 years, 2 months ago

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At the time I posted my comment, it said

For any gGg \in G and m,nZm,n \in \mathbb{Z}, then gm+n=gmgn=(gm)ng^{m+n} = g^mg^n = (g^m)^n and (gm)n=gmn\left( g^m \right)^n = g^{mn}.

I remember the strange height difference of the exponents, due to the use of normal brackets as opposed to \left( and \right).

Tim Vermeulen - 6 years, 2 months ago

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"The order of a group G, is the number of elements in G, which we denote by ∣G∣ like in set notation." This holds only if G is finite

Gabriel Romon - 6 years, 2 months ago

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Not necessarily. The post clearly gives examples of groups of infinite cardinality. A group is just like a set, but with a special metric on the set.

Bob Krueger - 6 years, 2 months ago

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a special metric? please clarify

o b - 6 years, 2 months ago

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Wait; for the extension problem (see below) the problem tells us that GG is cyclic, so the smallest gg such that gn=eg^n=e would be 22 right?

Sunay Joshi - 6 years, 2 months ago

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