# 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
5 years 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.

- 5 years ago

We can see it more easily as follows.

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

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

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

- 5 years 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?

- 5 years 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

- 5 years 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.

- 5 years ago

- 5 years ago

It says that

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

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

- 5 years 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 - 5 years 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).

- 5 years ago