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I've written an answer for an exercise in Artin's algebra. Can someone please verify it?

Which of the following groups are subgroups?

(a) $GL_n(\mathbb{R}) \subset GL_n(\mathbb{C})$

(b) $\{-1,1\} \subset \mathbb{R}^\times$

(c) The set of positive integers in $\mathbb{Z}^+$

(d) The set of positive reals in $\mathbb{R}^\times$

(e) The set of all matrices $\begin{bmatrix} a & 0 \\ 0 & 0 \end{bmatrix}$ in $GL_2(\mathbb{R})$

(a) $GL_n(\mathbb{R})$ is a subgroup of $GL_n(\mathbb{C})$, since it contains the identity matrix and is closed under matrix multiplication. Furthermore, each element has its inverse.

(b) $\{-1,1\}$ is a subgroup of $\mathbb{R}^\times$, since it contains the identity element $1$, is closed under multiplication. Furthermore, the inverse of $-1$ is $-1$, and that of $1$ is $1$.

(c) The set of positive integers in $\mathbb{Z}^+$ is not a subgroup, as it does not contain the identity element, $0$.

(d) The set of positive reals $H$ in $\mathbb{R}^\times$ is a subgroup since it is closed under multiplication, contains the identity element $1$, and each element $a \in H$ has the inverse $\frac{1}{a} \in H$.

(e) This is not a subgroup since it does not contain $I_2$, the $2 \times 2$ identity matrix.

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    $\begingroup$ Seems good to me. $\endgroup$ – Snufsan Jun 3 '14 at 8:31
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With (e) you must be careful: the matrices $$ \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} $$ are not invertible, hence are not elements of $GL_2(\mathbb R)$.

However, the set of matrices of this type $$ \left\{ \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} : \ a \in \mathbb R, \ a\ne 0 \right\} $$ is an Abelian group under matrix multiplication. Its identity is not the usual identity $I_2$...

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