Suppose G is a group of order 4 and $x^2=e$ for all $x$ in G. Prove that G is isomorphic to $Z_2\oplus Z_2$.

Question

Suppose $G$ is a group of order $4$ and $x^2=e$ for all $x$ in $G$. Prove that $G$ is isomorphic to $Z_2\oplus Z_2$.

My attempt:

(1) Show that $G$ is abelian.

\begin{align*} \text{Take }x,y\in G.&\text{Then }xy\in G.\\ (xy)^2&=e\\ \text{Thus } (xy)^{-1}&=xy\\ \text{Also by socks-shoe property }(xy)^{-1}&=y^{-1}x^{-1}=yx\\\\\text{Thus we have }&xy=yx \text{ and $G$ is abelian}. \end{align*} (2) Show that $Z_2\oplus Z_2$ is abelian. $$\text{This can be easily verified.}$$ (3) Define the isomorphism $\phi$ $$\text{Let $G$}=\{e,a,b,c\}\\ \text{Define } \phi \text{ from $G$ to $Z_2\oplus Z_2$ such that}\\ \phi(e)=(0,0)\\\phi(a)=(0,1)\\\phi(b)=(1,0)\\\phi(c)=(1,1)$$ $$\text{Clearly $\phi$ is one-one and onto.}$$

How do I show that $\phi$ is operation preserving?
If this is not the right way the proof is going, please post the solutions.

You need to show that $ab=c$, $bc=a$ and $ac=b$. All of which are the same. To show $ab=c$: Show that $ab$ cannot be one of the other three elements of the group. — David C. Ullrich, May 22 '16 at 15:17
Are you trying to say that i should show that $ab=a$,$ab=b$,$ab=e$ would all lead to $a=b$? — Bijesh K.S, May 22 '16 at 15:19
If you showed that that would do it. No, that's not what I was saying - was saying just that you need to show that $ab\ne a$, $ab\ne b$ and $ab\ne e$. — David C. Ullrich, May 22 '16 at 15:21
OH..So..if $ab=a$ then $b=e$ and if $ab=e$ then $b=a$, a contradiction. Isn't it? @DavidC.Ullrich — Bijesh K.S, May 22 '16 at 15:26

score 3 · Accepted Answer · edited Jun 12 '20 at 10:38

Claim. The elements of $G$ can be written as $\{e,a,b,ab\}$.

Proof. Since $G$ is a group the identity element $e\in G$. Since $G$ is a group of order $4$, there must exist another three non-identity elements which are also mutually distinct. Let them be $a,b$ and $c$. We claim that $ab=c$.

If not then $e,a,b,ab$ and $c$ all are elements of $G$. Furthermore (observe that $ab$ is not equal to any one of $a,b$ and $e$) all of them are mutually distinct. This contradicts the fact that the order of $G$ is $4$. So we are done.

Now define a homomorphism $\phi:G\to\mathbb{Z}_2\oplus \mathbb{Z}_2$ by $$\phi(e)=(0,0)$$$$\phi(a)=(0,1)$$$$\phi(b)=(1,0)$$I think that you can take it from here.

score 1 · Answer 2 · answered May 22 '16 at 16:09

1

You can easily show the group contains two normal subgroups of size $2$. There is a criterion that states that if $K$ and $H$ are normal subgroups of $G$ with $K\cap H =\{e\}$ and $KH=G$ then $G\cong K \times H$.

answered May 22 '16 at 16:09

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Suppose G is a group of order 4 and $x^2=e$ for all $x$ in G. Prove that G is isomorphic to $Z_2\oplus Z_2$.

2 Answers2