Good day! I am struggling with the following question in my group theory class for quite some time and would love to recieve a hint:
Let $\mathbb{F}$ denote some field and $\mathbb{F^*} = \mathbb{F} \setminus \{0\}$. Let us define the following sets: \begin{gather*} Q = \left\{ \begin{pmatrix} \alpha & \beta \\ 0 & 1 \end{pmatrix} : \: \alpha \in \mathbb{F^*}, \beta \in \mathbb{F} \right\} \\ P = \left\{ \begin{pmatrix} \alpha & \beta \\ 0 & \alpha^{-1} \end{pmatrix} : \: \alpha \in \mathbb{F^*}, \beta \in \mathbb{F} \right\} \end{gather*} Show that both $P$ and $Q$ are groups with regard to matrix multiplication and that they are isomorphic.
The first part of the question (showing that these are in facts groups) may be a little bit tricky because of the notation, specifically that $\alpha^{-1}$ denotes the multiplicative inverse as opposed to the additive inverse.
However, once understood, it is quite technical to show both are groups for example by showing these are subgroups of $GL_2(\mathbb{F})$.
Showing that these are isomorphic is the part of the question I had to think about for quite some time without any progress really. Firstly, I tried to construct a group isomorphism directly (naively), simply by defining: \begin{gather*} \varphi: P \rightarrow Q \\ \begin{pmatrix} \alpha & \beta \\ 0 & \alpha^{-1} \end{pmatrix} \rightarrow \begin{pmatrix} \alpha & \beta \\ 0 & 1 \end{pmatrix} \end{gather*} Quite expectedly, it doesn't work, as this map is not homomorphic. One can easily verify by taking two generic elements of $P$ (I add the calculation at the end as it's probably not too relevant but maybe it will wake some ideas up). I tried some other maps that seemed natural, but wasn't able to find an isomorphism.
I thought maybe a better direction would be to use some isomorphism theorem but can't really see how it would be possible to apply one here.
I also tried to find some groups that both $Q$ and $P$ are naturally isomorphic too, but wasn't really able to find one - I think if that was the case I would be able to find a direct isomorphism between them.
Any hints regarding this problem would be extremely appreciated! (costed me some hours of sleep this one :D) Thank you so much and have a lovely day!
A calculation that shows why $\varphi$ is not a homomorphism:
For some $A = \begin{pmatrix} \alpha_1 & \beta_1 \\ 0 & \alpha_1^{-1} \end{pmatrix}$, $B = \begin{pmatrix} \alpha_2 & \beta_2 \\ 0 & \alpha_2^{-1} \end{pmatrix} \in P$: \begin{gather*} \varphi(AB) = \begin{pmatrix} \alpha_1 \alpha_2 & \alpha_1 \beta_2 + \beta_1 \alpha_2^{-1} \\ 0 & 1 \end{pmatrix} \\ \varphi(A)\varphi(B) = \begin{pmatrix} \alpha_1 \alpha_2 & \alpha_1 \beta_2 + \beta_1 \\ 0 & 1 \end{pmatrix} \end{gather*}
Assuming $\text{char }\mathbb F \neq 2$:
$P$ has exactly 2 involutions:
$ \big\{-I,I\big\}$ because an involution has eigenvalues of $1$ and/or $-1$ but $\alpha=\alpha^{-1}$ so the diagonal is constant. Now use the fact that an involution must be diagonalizable (in $GL_n(\mathbb F)$) $\implies \beta =0$.
But $Q$ has at least 4 involutions, given by $\left\{ I, \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix},\begin{pmatrix} -1 & -1 \\ 0 & 1 \end{pmatrix}\right\}$
conclude: any homomorphism $\psi: Q\longrightarrow P$ cannot be injective, contradicting the existence of an isormoprhism $\varphi: P\longrightarrow Q$.