Show that any non-cyclic group of order $9$ is isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_3$

Question

Show that any non-cyclic group of order $9$ is isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_3$

The basic idea to show this is to take any non-cyclic group $G$ of order $9$ and apply the theorem below

Theorem: If $G$ is a group having two subgroups $M$ and $N$ such that

• $M$ and $N$ are normal in $G$
• $M \cap N = \{1_G\}$
• $G = MN$

Then $G \cong M \times N$

So now by Cauchy's theorem there exists an $a \in G$ such that $o(a) = 3$. In fact there are $8$ elements in $G$ of order $3$. If we choose $y \in G$ such that $y \not\in \langle a \rangle$ then $\langle y \rangle \cap \langle a \rangle = \{1_G\}$, and it then follows that $G = \langle a \rangle \langle y \rangle $ (I have skipped a few details here just to get to my question quicker)

So now all I have left to do is show that both $\langle a \rangle$ and $\langle y \rangle$ are normal in $G$, but I'm not really sure how to go about showing that. If I choose any $g \in G$, then I don't think that it follows immediately that $g\langle a \rangle g^{-1} \subseteq \langle a \rangle$ and $g\langle y \rangle g^{-1} \subseteq \langle y \rangle$.

How can I prove that $\langle a \rangle$ and $\langle y \rangle$ are normal in $G$?

Answer 1

Are you familiar with the Sylow theorems? As first Sylow theorem tells you that in a group of 9 elements you got to have an element a of order 3. Now, $<a>$ is a subgroup of index 3 in a group of order 9, which is a greatest prime divisor of 9, hence $<a>$ is normal in $G$. Clearly, $G/<a>$ is a group with three elements, so exactly $\mathbb{Z}_3$.

Another proof: Any group of order $p^2$ for $p$ prime is Abelian. Then just use the classification theorem for Abelian groups.

Third proof: So, if you got a group of 9 elements, you must have exactly one element of order 1, that is neutral. For all other elements, their order divides 9, Lagrange theorem. Now, if we have element of order 9, that element generates the whole group, so we stay with the case where all $a \neq e$ where $e$ is a neutral have order 3. Choose one such $a$. Group $<a>$ has three elements. You have $G - <a>$ has 6 elements. Choose $b \in G-<a>$. Now , $<b>$ is a subgroup of $G$ and $<a>$ is a subgroup of $G$. As $<a> \bigcap <b> = {e}$, the smallest subgroup of $G$ that contains $<a>$ and $<b>$ has more than three elements. Hence, it has to be $G$. So, $G = <a,b> = <a> \times <b>$ which is exactly $\mathbb{Z}_3 \times \mathbb{Z}_3$. Basically, this is the proof that works for all prime $p$.

Answer 2

A group $G$ of order $p^2$ is abelian, and hence every subgroup of $G$ is normal. If $G$ is non-cyclic, then there are $p^2-1$ elements of order $p$, grouped into $p+1$ subgroups of order $p$. Take any two of these latter, say $\langle x\rangle$ and $\langle y\rangle$. Now, $\langle x\rangle\langle y\rangle$ is a subgroup of $G$ of order $p^2$, so it is the whole $G$, namely: $G=\langle x\rangle\langle y\rangle$.

Lemma. $\varphi\colon G\to \Bbb Z_p\oplus\Bbb Z_p$, defined by $x^iy^j\mapsto([i]_p,[j]_p)$ is an isomorphism.

Proof.

• Injectivity: $([i]_p,[j]_p)=([k]_p,[l]_p)\Longrightarrow$ $[i]_p=[k]_p\text{ and }$ $[j]_p=[l]_p\Longrightarrow$ $\exists m,n\in\Bbb Z:i=k+mp\text{ and }$ $j=l+np\Longrightarrow$ $x^i=x^{k+mp}=x^k\text{ and }$ $y^j=y^{l+mp}=y^l\Longrightarrow$ $x^iy^j=x^ky^l$.
• Surjectivity: by construction.
• Homomorphism: $\varphi((x^iy^j)(x^ky^l))\stackrel{G\text{ abelian}}{=}\varphi(x^{i+k\pmod p}y^{j+l\pmod p})=$ $([i+k]_p,[j+l]_p)=$ $([i]_p+[k]_p,[j]_p+[l]_p)=$ $([i]_p,[j]_p)+([k]_p,[l]_p)=$ $\varphi(x^iy^j)+\varphi(x^ky^l)$. $\space\Box$

Therefore, $G\cong\Bbb Z_p\oplus\Bbb Z_p$.