Exercise 2.5.10 (A group of order 4 is isomorphic to $Z_4$ or $Z_2 imes Z_2$)

Question

Classify groups of order $4$ by proving that if $|G| = 4$ then $G \simeq Z_4$ or $G \simeq V_4$.[See Exercise 36, Section 1.1.]

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} Let $G = \{1,a,b,c\}$ be a group of order $4$, where $e,a,b,c$ are distinct. By Lagrange's Theorem (see Ex. 1.7.19), for every $x \in G$, $|x|= |\langle x angle |$ is a divisor of $4$.

If there is some element $a\in G$ of order $4$, then $G = \langle a angle$ is cyclic, isomorphic to $Z_4$, by the isomorphism $Z_4 \mapsto G$ defined by $x^n \mapsto a^n$.

If not, every element $x 
e 1$ has order $2$, therefore
$$\forall x \in G,\ x^2 = 1.$$
This implies that $G$ is abelian: indeed, if $x, y \in G$, then $1 = (xy)^2 = xy xy$ thus $xy = y^{-1}x^{-1}$, and $x^{-1}= x$, $y^{-1}= y$, so $xy = yx$.
Consider the map
$$\varphi\ 
\left\{
\begin{array}{ccl}
\Z/2\Z 	imes \Z/2\Z & 	o & G\
(\overline{i},\overline{j}) & \mapsto & a^i b^j
\end{array}
ight.
$$
Then
\begin{itemize}
\item $\varphi$ is well defined:

If $i \equiv i' \pmod 2$ and $j \equiv j' \pmod 2$, then $a^i = a^{i'}$ and $b^j = b^{j'}$ because $|a| = |b| = 2$. Therefore $a^i b^j = a^{i'}b^{j'}$.

\item $\varphi$ is a homomorphism: if $(\overline{i},\overline{j})  \in \Z/2\Z 	imes \Z/2\Z$ and $(\overline{k},\overline{l})  \in \Z/2\Z 	imes \Z/2\Z$, then, since $G$ is abelian,
\begin{align*}
\varphi[(\overline{i},\overline{j})+ (\overline{k},\overline{l}) ]&= \varphi( (\overline{i +k},\overline{j+l})\
&= a^{i+k} b^{j+l}\
&= (a^i  b^j) (a^k b^l)\
&= \varphi(\overline{i},\overline{j}) \varphi(\overline{k},\overline{l}).
\end{align*}
\item $\varphi$ is injective: If $(\overline{i},\overline{j}) \in \ker(\varphi)$, then $a^i b^j = 1$. Therefore $a^i = b^{-j}$. If $\overline{i} = \overline{1}$, then $a \in \langle b angle = \{e, b\}$. This is impossible, because $a
e e$ and $a 
e b$, so $\overline{i} = \overline{0}$. This gives $1 = b^j$, where $1 
e b$, thus $j$ is even and $\overline{j} = \overline{0}$. Hence $\ker(\varphi) = \{(\overline{0}, \overline{0})\}$, so $\varphi$ is injective.
\item Since $\varphi : \Z/2\Z 	imes \Z/2\Z 	o  G$ is injective, where $|\Z/2\Z 	imes \Z/2\Z| = |G| = 4$, then $\varphi$ is surjective.
\end{itemize}
This shows that $\varphi$ is an isomorphism, so
$$G \simeq \Z/2\Z 	imes \Z/2\Z \simeq Z_2 	imes Z_2.$$

In conclusion, if $|G| = 4$ then $G \simeq Z_4$ or $G \simeq Z_2 	imes Z_2$.

\bigskip
(If we name $V_4$ a non cyclic group of order $4$ (Klein's {\bf V}ierergruppe), then $Z_2 	imes Z_2 \simeq V_4$ : see the table in Ex. 1.1.36.)
\end{proof}

Exercise 2.5.10 (A group of order 4 is isomorphic to $Z_4$ or $Z_2\times Z_2$)

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Exercise 2.5.10 (A group of order 4 is isomorphic to $Z_4$ or $Z_2\times Z_2$)

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