Let the group $G$ be defined by the generators $x,y$ and the relations $x^8=y^2=yxyx^5=e$, where $e$ is the neutral element, i.e.:
$$
G= \left<x,y \ | \ x^8=y^2=yxyx^5=e \right>
$$
I have to determine if $G$ is finite and provide a list of elements. Now by considering the last equality and using that $y^{-1}=y$:
$$
y^{-1}xy=x^{-5} \in \left< x \right>
$$
And thus $\left< x \right>$ is normal in $G$, furthermore we know that $\left< x \right> \leq C_8$, $\left< y \right> \leq C_2$, with $C_n$ the cyclic group of order $n$. This makes me think that $G \cong C_8 \times C_2$. How can I make this more rigorous/prove my suspicion?
Best Answer
The presentation that you have presented can be re-written as $$ \langle x, y \mid x^8 = y^2 = e, y x y^{-1} = x^3\rangle, $$ which is a canonical presentation of $G$ as an inner semidirect product $\langle x\rangle \rtimes \langle y\rangle$.
To be more explicit, note that since $\langle x \rangle$ is normal, any element of $G$ can be written as $x^a y^b$ for some $a \in C_8, b \in C_2$ (you should show this). We can then compute the product of two elements $x^a y^b \cdot x^{a'} y^{b'}$. If $b = 0$, we have $x^a y^b \cdot x^{a'} y^{b'} = x^{a+a'}y^{b'} = x^{a+a'}y^{b + b'}$, whereas if $b=1$, we can use the conjugation relation to see that $yx = x^3y$, and so $$ x^a y \cdot x^{a'}y^{b'} = x^a (x^3 y) x^{a'-1} y^{b'} = \dotsb = x^a (x^{3a'} y) y^{b'} = x^{a + 3a'} y^{b + b'}. $$
Putting these together, we can write $x^a y^b \cdot x^{a'}y^{b'} = x^{a + (2b+1)a'} y^{b+ b'}$.
Let $\varphi\colon C_2 \rightarrow\mathrm{Aut}(C_8)$ be defined by $\varphi(b) \colon a \mapsto (2b + 1)a$ (this is an automorphism since $3$ and $8$ are coprime), which is a homomorphism and satisfies $x^a y^b \cdot x^{a'}y^{b'} = x^{a + \varphi(b)(a')} y^{b+ b'}$. But this exactly means that the map \begin{align*} \lambda \colon\; G &\rightarrow C_8 \rtimes_\varphi C_2 \\ x^a y^b &\mapsto (a, b) \end{align*} is a homomorphism and so, since each $x^a y^b$ is distinct for $a \in C_8, b \in C_2$, an isomorphism. That is, $G \cong C_8 \rtimes_\varphi C_2$.