I'm stuck on the following question:
It is given that \begin{equation*} G = \left\{\begin{pmatrix}\bar{a}&\bar{b}\\ \bar{0}&\bar{c}\end{pmatrix}\text{ with $\bar{a}$ and $\bar{c}$ in $\mathbb{F}_7^*$ and $\bar{b}\in\mathbb{F}_7$}\right\} \end{equation*} is a subgroup of $GL_2(\mathbb{F}_7)$.
- Determine which elements of $G$ have order 2.
- Compute the center $Z(G)$ of $G$
For part one I found that $$\begin{pmatrix}\bar{a}&\bar{b}\\ \bar{0}&\bar{c}\end{pmatrix}^2 = \begin{pmatrix}\overline{a^2}&\overline{b(a+c)}\\ \bar{0}&\overline{c^2}\end{pmatrix}$$ so we must have $\bar{a}=\bar{c}=\bar{1}$ and $\bar{b}=\bar{0}$. But that's just the identity, so is there no element of order 2 or am I missing something? For the second part I'm not really sure where to start.
To calculate $Z(G)$, we can use $$ Z(G)=\bigcap_{g\in G} C_G(g) \leqslant \bigcap_{g\in G} C_{GL(2,7)}(g) \leqslant C_{GL(2,7)}(g_1)\cap C_{GL(2,7)}(g_2)\cap C_{GL(2,7)}(g_3) $$ where $$ g_1=\pmatrix{1 & 0\\0 &-1},\ g_2=\pmatrix{-1 & 0\\0 &1},\ g_3=\pmatrix{1 & 1\\0 &1}. $$
To simplify the calculation even further, note that any matrix commuting with these will commute with $$ 4(g_1 +I)=\pmatrix{1 & 0\\0 & 0}=e_{11},\ 4(g_2 +I)=\pmatrix{0 & 0\\0 & 1}=e_{22},\ g_3 - I=\pmatrix{0 & 1\\0 & 0}=e_{12}. $$
It's immediate to see that the matrices commuting with these three are just the scalars, so that $$Z(G)\leqslant \{ cI\mid c\in\mathbb{F}_{7}^{*}\};$$ and of course the reverse implication is even more clear.