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Let $\struct {G, \circ}$ be an algebraic structure.

$\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$ if and only if:

$(1): \quad \struct {H, \circ}$ is a group
$(2): \quad H$ is a subset of $G$.

This is represented symbolically as $H \le G$.

It is usual that $\struct {G, \circ}$ is itself a group, but that is not necessary for the definition.


$\N$ in $\struct {\R_{\ne 0}, \times}$

Consider the multiplicative group of real numbers $\struct {\R_{\ne 0}, \times}$.

Consider the algebraic structure $\struct {\N_{> 0}, \times}$ formed by the non-zero natural numbers under multiplication.

Then $\struct {\N_{> 0}, \times}$ is not a subgroup of $\struct {\R_{\ne 0}, \times}$.

Matrices $\begin{bmatrix} 1 & a \cr 0 & 1 \end{bmatrix}$ in General Linear Group

Let $\GL 2$ denote the general linear group of order $2$.

Let $H$ be the set of square matrices of the form $\begin{bmatrix} 1 & a \cr 0 & 1 \end{bmatrix}$ for $a \in \R$.

Then $\struct {H, \times}$ is a subgroup of $\GL 2$, where $\times$ is used to denote (conventional) matrix multiplication.

Also see

  • Results about subgroups can be found here.