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.
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 \(H\not\leq G\text{.}\)
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 \(G\text{.}\)
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).
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 \(\Q\text{.}\)
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 \(\R\text{.}\)
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 \(\R^+\text{.}\)
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 \(\M_n(\R)\text{.}\)
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 \(\Z_4\text{.}\)
Let \(G\) be a group. Then \(\{e_G\}\) and \(G\) are clearly both subgroups of \(G\text{.}\)
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 \(G\text{.}\)
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.
Let \(G\) be a group.
If \(H\) is a subgroup of \(G\) then the identity element \(e_H\) of \(H\) is \(e_G\text{,}\) the identity element of \(G\text{.}\)
If \(H\) is a subgroup of \(G\) and \(a\in H\) has inverse \(a^{-1}\) in \(G\text{,}\) then \(a\)'s inverse in \(H\) is also \(a^{-1}\text{.}\)
Let \(H\subseteq G\text{.}\) If the identity element of \(G\) is not in \(H\text{,}\) then \(H\not\leq G\text{.}\)