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Group Theory

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

How would you prove the following?

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

Note by Calvin Lin
4 years, 3 months ago

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Let \(a,b \in G\). Note that \((ab)^2 = e\) because \(ab \in G\). So \[e = (ab)^2 = abab\] Hence \[aeb = a(abab)b\] \[ab = (aa)(ba)(bb)\] \[ab = a^2bab^2\] \[ab = ebae\] \[ab = ba\] for all \(a,b \in G\), that is to say \(G\) is abelian.

Jan J. - 4 years, 3 months ago

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We can see it more easily as follows.

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

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

Abhishek Sinha - 4 years, 3 months ago

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

Samuel Queen - 4 years, 3 months ago

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Wait; for the extension problem (see below) the problem tells us that \(G\) is cyclic, so the smallest \(g\) such that \(g^n=e\) would be \(2\) right?

Sunay Joshi - 4 years, 3 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 - 4 years, 3 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 - 4 years, 3 months ago

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a special metric? please clarify

O B - 4 years, 3 months ago

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It says that

\[ g^{m+n} = g^mg^n = (g^m)^n, \]

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

Tim Vermeulen - 4 years, 3 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 \(g \in G\) and \(m, n \in \mathbb{Z} \), then \( g^{m+n} = g^m g^n \) and \( \left( g^m \right)^n =g^{mn} \).

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

Calvin Lin Staff - 4 years, 3 months ago

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At the time I posted my comment, it said

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

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

Tim Vermeulen - 4 years, 3 months ago

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