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Suppose we have the following: $$G=\Bigg\{ \left[ \begin{array}{ c c } \overline{a} & \overline{b} \\ \overline{0} & \overline{c} \end{array} \right] \Bigg|\overline{a},\overline{c} \in \mathbb{F}_3^* \text{ and } \overline{b} \in \mathbb{F}_3 \Bigg\} $$

$$H=\Bigg\{ \left[ \begin{array}{ c c } \overline{0} & \overline{b} \\ \overline{c} & \overline{0} \end{array} \right] \Bigg| \overline{b},\overline{c} \in \mathbb{F}_3^* \Bigg\} $$.

$G$ is obviously a subgroup of $GL_2(\mathbb{F}_3)$ and the identity element is $ \left[ \begin{array}{ c c } \overline{1} & \overline{0} \\ \overline{0} & \overline{1} \end{array} \right]$.

My question is: is $H$ a group? Since $b,c \in \mathbb{F}_3^*=\{\overline{1},\overline{2}\}$, the identity matrix of $2 \times 2$-matrices isn't an element of $H$ because $b$ and $c$ can't equal $\overline{0}$ and the other coefficients are always $\overline{0}$ and thus can't equal $\overline{1}$. So is there some other identity for $H$? Or is it impossible for $H$ to be a group because of this?

Also, for all $GL_n(F)$ the identity element is just the $n \times n$ identity matrix, right? I'm a little thrown off by the integers modulo $n$ thing because some matrix coefficients can come from a group under addition and others from a group under multiplication.

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    $\begingroup$ $H$ is not even closed, as $A := \pmatrix{1&0\\1&1} \in H$ but $A^3 = I \not\in H$. $\endgroup$ – Travis Willse Jun 7 '16 at 13:35
  • $\begingroup$ Typo in your description of $H$? $\endgroup$ – D_S Jun 7 '16 at 13:36
  • $\begingroup$ Indeed, $H$ as described is not a group $\endgroup$ – Omnomnomnom Jun 7 '16 at 13:40
  • $\begingroup$ I just made up some $H$ without checking if it's closed under multiplication and inverses. I've edited $H$ now so that it's closed. $\endgroup$ – Clark Jun 7 '16 at 13:43
  • $\begingroup$ So if I understand correctly $H$ is still not a group because it doesn't have an identity element? $\endgroup$ – Clark Jun 7 '16 at 13:53
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in a group, the identity element is unique, and for every element $x$ the inverse $x^{-1}$ of $x$ also is unique.

As $H$ does not contain the identity element $I_2$, it is not a subgroup of $GL_2 (F_3)$. Clearly $I_n$ is the neutral element of $GL_n (F)$ and is unique.

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