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Section4.1Introduction to subgroups

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.

Proof