Let $H$ be an order 9 group generated by 2 elements $a,b$ that commute and have at most order 3.

Question

Let $H$ be a group with identity $1_H$ that is generated by 2 elements $a,b$ that commute ($ab=ba$) and where each has at most order $3$. In symbols (I hope I translated correctly):

$$H=\langle a,b\rangle \ \text{, where} \ a^3=b^3=1_H=a^{-1}b^{-1}ab$$

Assuming $H$ has exactly order 9 and assuming $\{a,b,1_H\}$ are all distinct, what is $H$ isomorphic to? (order 9 possibilities are $\mathbb Z_9$ and $\mathbb Z_3 \times \mathbb Z_3$)

---

For order 9: Assuming $H$ is of order 9, I believe $H$ is isomorphic to $\mathbb Z_3 \times \mathbb Z_3$.

Construct a map $\gamma: \mathbb Z_3^2 \to H$, $\gamma(c \times d)=b^ca^d$, where $c,d \in \{0,1,2\}$.

Show $\gamma$ is bijective: obvious

Show $\gamma$ is a homomorphism: For each $c,d,e,f \in \{0,1,2\}$, we must show that

(notation: instead of $(c,d) \in \mathbb Z_3^2$, I'll say $c \times d$)

$$\gamma(c \times d + e \times f) = \gamma(c \times d) \gamma(e \times f)$$

I believe this is equivalent to

$$b^{c+d}a^{e+f} = b^ca^d b^ea^f$$

Finally, because $a^db^e=b^ea^d$ for $d,e \in \{0,1,2\}$, we have that

$$RHS = b^ca^d b^ea^f = b^cb^e a^da^f = LHS$$

Answer 1

You don't have to go back that far.

The possibilities of $H$ is $\mu_9$ or $\mu_3\times\mu_3$. The former has an element of order 9(the generator), and the latter has no element of order 9.

Every element of $H$ is looks like this: $a^mb^n$, and it has order 3 or less, so it can't be order 9(because $(a^mb^n)^3=a^mb^na^mb^na^mb^n=a^{3m}b^{3n}=(a^3)^m(b^3)^n=1_H^m1_H^n=1_H$.)

So $H\cong\mu_3\times\mu_3$ .

Answer 2

Notation: Elements of $X\times Y$ are generally written as ordered pairs $(x,y)$, not $x\times y$. Also, avoid using $e$ and an element, as its usually reserved for additive identities.

To be clear, we are looking at $\varphi:(\mathbb Z_3\times \mathbb Z_3,+)\to (H,\cdot)$ defined by $\varphi((c,d))=b^c\cdot a^d$. Therefore,

\begin{align*} \varphi((x,y)+(z,w))&=\varphi((x+z,y+w))\\ &=b^{x+z}\cdot a^{y+w}\\ &=b^x\cdot b^z\cdot a^y\cdot a^w\\ &=(b^x\cdot a^y)\cdot (b^z\cdot a^w)\\ &=\varphi((x,y))\cdot\varphi((z,w)), \end{align*}

so it is indeed a homomorphism.