Abelian group

[Tolaso J Kos]

Prove that a group $(G, *)$ that has either $1$ or $2$ or $3$ or $4$ elements is abelian.

[Papapetros Vaggelis]

Hi Tolaso.

Let \(\displaystyle{\left(G,\cdot\right)}\) be a group.

If \(\displaystyle{|G|=1}\), then \(\displaystyle{G=\left\{e\right\}}\) and it's ok.

If \(\displaystyle{|G|=2\,\,,G=\left\{e,a\right\}}\), then \(\displaystyle{\left(G,\cdot\right)}\) is abelian

and \(\displaystyle{\left(G,\cdot\right)\simeq \left(\mathbb{Z}_{2},+\right)}\) .

Let \(\displaystyle{|G|=3\,\,,G=\left\{e,a,b\right\}}\).

1st proof

We have that \(\displaystyle{e\,a=a\,e=a\,\,,e\,b=b\,e}\). If \(\displaystyle{a\,b=a}\) or \(\displaystyle{a\,b=b}\)

then \(\displaystyle{b=e}\) or \(\displaystyle{a=e}\), a contradiction. So,

\(\displaystyle{a\,b=e=b\,a}\) and \(\displaystyle{\left(G,\cdot\right)}\) is abelian.

2nd proof

If \(\displaystyle{H}\) is a subgroup of \(\displaystyle{\left(G,\cdot\right)}\), then

\(\displaystyle{|H|\mid |G|=3\implies |H|\in\left\{1,3\right\}\implies H\in\left\{\left\{e\right\},G\right\}}\).

So, \(\displaystyle{\forall\,x\in G-\left\{e\right\}, G=\langle{x\rangle}}\) and \(\displaystyle{\left(G,\cdot\right)}\)

is abelian as cyclic and thus \(\displaystyle{\left(G,\cdot\right)\simeq \left(\mathbb{Z}_{3},+\right)}\) .

Last case

Let \(\displaystyle{|G|=4=2^2}\). If \(\displaystyle{x\,,y\in G}\), then

\(\displaystyle{|\langle{x\rangle}|\,,|\langle{y\rangle}|\in\left\{1,2,4\right\}}\) .

If \(\displaystyle{|\langle{x\rangle}|=1}\) or \(\displaystyle{|\langle{y\rangle}|=1}\), then

\(\displaystyle{x=e}\) or \(\displaystyle{y=e}\) and \(\displaystyle{x\,y=y\,x\in\left\{x,y\right\}}\) .

If \(\displaystyle{|\langle{x\rangle}|=4}\) or \(\displaystyle{|\langle{y\rangle}|=4}\) , then

\(\displaystyle{G=\langle{x\rangle}}\), which means that \(\displaystyle{G}\) is abelian and

\(\displaystyle{\left(G,\cdot\right)\simeq \left(\mathbb{Z}_{4},+\right)}\) .

Let \(\displaystyle{|\langle{x\rangle}|=2=|\langle{y\rangle}|}\) . Then,

\(\displaystyle{\langle{x\rangle}\cap \langle{y\rangle}\leq \langle{x\rangle}}\), so, if

\(\displaystyle{\langle{x\rangle}\cap \langle{y\rangle}=\langle{x\rangle}\iff \langle{x\rangle}\subseteq \langle{y\rangle}}\)

and then \(\displaystyle{x\,y=y\,x}\). If \(\displaystyle{\langle{x\rangle}\cap \langle{y\rangle}=\left\{e\right\}}\), then

since \(\displaystyle{\langle{x\rangle}\,,\langle{y\rangle}}\) are normal subgroups, we get

\(\displaystyle{G=\langle{x\rangle}\cdot \langle{y\rangle}\simeq \mathbb{Z}_{2}\times \mathbb{Z}_{2}}\) .

Note

The usual way is to make the board of multiplication. However, the above solution, gives us also

all the groups of order \(\displaystyle{1,2,3}\) and \(\displaystyle{4}\) .