Sometimes groups are too complicated to understand directly. One method that can be used to identify a group's structure is to study its subgroups.

##### Definition4.1.1

A subgroup of a group $G$ is a subset of $G$ that is also a group under $G$'s operation. If $H$ is a subgroup of $G\text{,}$ we write $H \leq G\text{;}$ if $H\subseteq G$ is not a subgroup of $G\text{,}$ we write $H\not\leq G\text{.}$

##### Warning4.1.2

Do not confuse subgroups with subsets! All subgroups of a group $G$ are, by definition, subsets of $G\text{,}$ but not all subsets of $G$ are subgroups of $G$ (see Example 4.1.3, Parts 2–5, below). Whether or not a subset of $G$ is a subgroup of $G$ depends on the operation of $G\text{.}$

##### Example4.1.3

1. Consider the subset $\Z$ of the group $\Q\text{,}$ assuming that $\Q$ is equipped with the usual addition of real numbers (as we indicated above that we would assume, by default). Since we already know that $\Z$ is a group under this operation, $\Z$ is not just a subset but in fact a subgroup of $\Q$ (under addition).

2. Instead, consider the subset $\Q^+$ of the group $\Q\text{.}$ This subset is not a group under $\Q$'s operation $+\text{,}$ since it does not contain an identity element for $+\text{.}$ Therefore, $\Q^+$ is a subset but not a subgroup of $\Q\text{.}$

3. Let $I$ be the subset

\begin{equation*} I=\R-\Q=\{x\in \R: x \text{ is irrational} \} \end{equation*}

of the group $\R\text{.}$ The set $I$ is not a group under $\R$'s operation $+$ since it is not closed under addition: for instance, $\pi, -\pi \in I\text{,}$ but $\pi+(-\pi)=0\not\in I\text{.}$ So $I$ is a subset but not a subgroup of $\R\text{.}$

4. Consider the subset $\Z^+$ of the group $\R^+\text{.}$ The set $\Z^+$ is closed under multiplication, multiplication is associative on $\Z^+\text{,}$ and $\Z^+$ does contain an identity element (namely, 1). However, most elements of $\Z^+$ do not have inverses in $\Z^+$ under multiplication: for instance, the inverse of $3$ would have to be $1/3\text{,}$ but $1/3\not\in \Z^+\text{.}$ Therefore, $\Z^+$ is a subset but not a subgroup of $\R^+\text{.}$

5. Consider the subset $GL(n,\R)$ of $\M_n(\R)\text{.}$ We know that $GL(n,\R)$ is a group, so it might be tempting to say that it is a subgroup of $\M_n(\R)\text{;}$ to be a subgroup of $\M_n(\R)\text{,}$ $GL(n,\R)$ must be a group under $\M_n(\R)$'s operation, which is matrix addition. While $GL(n,\R)$ is a group under matrix multiplication, it is not a group under matrix addition: for instance, it is not closed under matrix addition, since $I_n, -I_n\in GL(n,\R)$ but $I_n+(-I_n)$ is the matrix consisting of all zeros, which is not in $GL(n,\R)\text{.}$ So $GL(n,\R)$ is a subset but not a subgroup of $\M_n(\R)\text{.}$

6. Consider the subset $H=\{0,2\}$ of $\Z_4\text{.}$ The subset $H$ is closed under addition modulo 4 ($0+0=0\text{,}$ $0+2=2+0=2\text{,}$ $2+2=0$), addition modulo 4 is always associative, $H$ contains an identity element (namely, 0) under addition modulo 4, and both 0 and 2 have inverses in $\Z_4$ under this operation (0 and 2 are each their own inverses). Thus, $H$ is a subgroup of $\Z_4\text{.}$

7. Let $G$ be a group. Then $\{e_G\}$ and $G$ are clearly both subgroups of $G\text{.}$

##### Definition4.1.4

Let $G$ be a group. The subgroups $\{e_G\}$ and $G$ of $G$ are called the trivial subgroup and the improper subgroup of $G\text{,}$ respectively. Not surprisingly, if $H\leq G$ and $H\neq \{e_G\}\text{,}$ $H$ is called a nontrivial subgroup of $G\text{,}$ and if $H\leq G$ and $H\neq G\text{,}$ $H$ is called a proper subgroup of $G\text{.}$

##### Remark4.1.5

Sometimes the notation $H\lt G$ is used to indicate that $H$ is a proper subgroup of $G\text{,}$ but sometimes it is simply used to mean that $H$ is a subgroup—proper or improper—of $G\text{.}$ We will not use the notation $H\lt G$ in this text.

Notice that in the cases above, we saw subsets of groups fail to be subgroups because they were not closed under the groups' operations; because they did not contain identity elements; or because they didn't contain an inverse for each of their elements. None, however, failed because the relevant group's operation was not associative on them. This is not a coincidence: rather, since any element of a subset of a group $G$ also lives in $G\text{,}$ any associative operation on $G$ is of necessity associative on any closed subset of $G\text{.}$ Therefore, when we are checking to see if $H\subseteq G$ is a subgroup of group $G\text{,}$ we need only check for closure, an identity element, and inverses.