Show that two finite two generator groups are isomorphic

Question

I was faced by a question that I can't solve. Any help would be great! Let $A$ and $B$ be groups with the following properties:

\begin{cases} |A| = 9 \cdot 3=27\\ A = \left<a,b\right> \\ a^{9} = e \\ b^{3} = e \\ ba = a^{4}b \end{cases}

\begin{cases} |B| = 9 \cdot3=27\\ B = \left<\alpha,\beta\right> \\ \alpha^{9} = e \\ \beta^{3} = e \\ \beta\alpha = \alpha^{7}\beta \end{cases} I need to prove that $A \cong B$.
I really can't think of any isomorphism (or how to use some theorem) to solve this. Thanks in advance!

Answer 1

Let’s drop the assumption that $A$ and $B$ have order $27$ (it is not needed). Consider the presentations: $$\begin{align*} A &= \langle a,b\mid a^9 = b^3=e,\ ba=a^4b\rangle\\ B &= \langle \alpha,\beta\mid \alpha^9=\beta^3=e,\ \beta\alpha=\alpha^7\beta\rangle. \end{align*}$$ By von Dyck’s Theorem, we get a morphism from $A$ to $B$ by identifying two elements of $B$ that satisfy the same relations as $a$ and $b$ do, in $B$; and we get a morphism from $B$ to $A$ by identifying two elements of $A$ that satisfy in $A$ the same relations that $\alpha$ and $\beta$ do in $B$.

Consider the pair $(\alpha,\beta^2)$ in $B$. I claim that they satisfy the same relations in $B$ that $a$ and $b$ do in $A$. Indeed, $\alpha^9 =e$, $(\beta^2)^3= (\beta^3)^2 = e^2 = e$. And $$\beta^2\alpha = \beta(\alpha^7\beta) = \alpha^{49}\beta^2 = \alpha\beta^2.$$ Thus, the map $a\mapsto \alpha$, $b\mapsto \beta^2$ extends uniquely to a morphism $A\to B$.

Now consider the elements $(a,b^2)$ in $A$. We claim they satisfyt the same relations as $\alpha$ and $\beta$ do in $B$. Indeed, we have $a^9 = e$, $(b^2)^3 = (b^3)^2 = e^2 = e$, and $$b^2a = b(a^4b) = a^{16}b^2 = a^7b^2.$$ Therefore, the map $\alpha\mapsto a$ and $\beta\mapsto b^2$ extends uniquely to a morphism $B\to A$.

Now note that the two morphisms are inverses of each other: we have $$\begin{align*} a&\mapsto \alpha\mapsto a&\quad b&\mapsto \beta^2\mapsto (b^2)^2 = b\\ \alpha&\mapsto a\mapsto \alpha &\quad \beta&\mapsto b^2\mapsto (\beta^2)^2 = \beta. \end{align*}$$ Thus, the two maps are isomorphisms, so $A\cong B$.

Now, we can show that $A$ and $B$ are in fact of order $27$. It’s easy to show $A$ has cardinality at most $27$ by showing every element can be written in the form $a^ib^j$, $0\leq i\lt 9$, $0 \leq j \lt 3$. And then we can consider the semidirect product $C_7\rtimes_{\phi} C_3$, with $C_7$ generated by $x$ and $C_3$ generated by $y$, and where $\phi\colon C_3\to \mathrm{Aut}(C_7)\cong C_6$ sends $y$ to the automorphism $x\mapsto x^4$. Then $x$ and $y$ in $C_7\rtimes_{\phi}C_3$ satisfy the same realtions as $a$ and $b$ to in $A$, so $a\mapsto x$ and $b\mapsto y$ extends to a homomorphism $A\to C_7\rtimes_{\phi} C_3$. Since $x$ and $y$ generate the latter, the morphism is surjective, so $|A|\geq 27$, proving the desired equality.

Answer 2

Thanks to Jeremy Rickard for his comments on my previous, incorrect, answer.

First, it's easy to prove by induction on $n$ that $ba^n=a^{4n}b$. It's then easy to prove by induction on $k$ that $b^ka^n=a^{{4^k}n}b^k.$ Similarly, $\beta^k\alpha^n=\alpha^{{7^k}n}\beta^k.$ Thus, we observe that $b^2a=a^{16}b^2=a^7b^2$. This observation motivates the following effort.

Define $\varphi:A \to B$ by $\varphi(a)=\alpha, \varphi(b)=\beta^2$. It's easy to confirm that $\varphi(a^9)=\varphi(b)^3=1_B$, that $\varphi(a)$ in fact has order $9$, and that $\varphi(b)$ in fact has order $3$. Also:

$$\varphi(ba)=\varphi(a^4b)=\varphi(a)^4\varphi(b)=\alpha^4\beta^2=\alpha^{49}\beta^2=\beta^2\alpha=\varphi(b)\varphi(a).$$

Thus, $\varphi$ is well-defined and a homomorphism, so $\varphi(A)$ is a subgroup of $B$, which meanns its order is a power of $3$ no greater than $27$. But $\alpha \in \varphi(A) \Rightarrow \vert \varphi(A) \vert \geq 9$ and $\beta^2 \in \varphi(A) \setminus \langle \alpha \rangle$, so in fact $\vert \varphi(A) \vert \gt 9$. It follows that $\varphi(A)=B$ and since $\vert A \vert = \vert B \vert, \varphi:A \to B$ is $1$-$1$, and hence is an isomorphism.