# 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
6 years, 5 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.

- 6 years, 5 months ago

We can see it more easily as follows.

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

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

- 6 years, 5 months ago

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 $g \in G \Longrightarrow g^{n} = e$.

- 6 years, 5 months ago

It says that

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

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

- 6 years, 5 months ago

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.

Staff - 6 years, 5 months ago

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).

- 6 years, 5 months ago

"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

- 6 years, 5 months ago

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.

- 6 years, 5 months ago

- 6 years, 5 months ago

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?

- 6 years, 5 months ago