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 \(H\not\leq G\text{.}\)
Example4.1.3
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 \(G\text{.}\)
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 \(G\text{.}\)
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.
Lemma4.1.6
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 \(a^{1}\text{.}\)
For Part 1: Since \(e_H\) is in both \(H\) and \(G\text{,}\) by the definition of \(e_H\text{,}\) we have \(e_He_H=e_H\text{,}\) and by the definition of \(e_G\) we have \(e_Ge_H=e_H\text{.}\) So \(e_He_H=e_Ge_H\text{,}\) and thus by right cancellation, \(e_H=e_G\text{.}\)
Next, for Part 2, let \(b\) be the inverse of \(a\) in \(H\text{.}\) Then using Part 1 of this lemma and the definition of an inverse, \(ab=e_H=e_G=aa^{1}\text{.}\) By left cancellation, then, we have that \(b=a^{1}\text{.}\)
Corollary4.1.7
Let \(H\subseteq G\text{.}\) If the identity element of \(G\) is not in \(H\text{,}\) then \(H\not\leq G\text{.}\)