## 4.6 Subgroups

Recall the group \(G=\{1, -1, i, -i\}, \times\), where *i* is a square
root of \(-1\) in \(\CC\). Consider the subset \(H = \{1,-1\}\) of *G*. Then
*H* is still a group with the same group operation as *G* – it is a
group sitting inside *G* with the same operation.

On the other hand \(K = \{1, i\} \subset G\) doesn’t form a group under
\(\times\). Indeed \(\times\) doesn’t even give a binary operation on *K*,
because \(i \times i = -1 \notin K\).

As another example, consider the group \(\ZZ, +\). Then \(2\ZZ\) which is
defined to be \(\{ 2z : z \in \ZZ\}\), that is the set of even integers,
is again a group under *+* (check this!). Now consider \(\mathbb{N} = \{0,1,2,\ldots\} \subset \ZZ\). This time *+* *does* give a binary
operation on \(\mathbb{N}\), but \(\mathbb{N}\) isn’t a group under *+*
(why?).

We capture the notion of a group inside another group with the following definition:

**Definition 4.11 **Let \((G, *)\) be a group with identity *e*. Let \(H \subseteq G\) be such that

- \(e \in H\),
- if \(h,k \in H\) then \(h*k \in H\), and
- if \(h \in H\) then \(h^{-1} \in H\).

*H*is a subgroup of

*G*and write \(H \leq G\).

The important part of the definition is that *H* has to be a group with
the same group operation as *G*.

**Example 4.5 **

- \(\{1, -1\}\) is a subgroup of \((\{1,-1,i, -i\}, \times)\) which is itself a subgroup of \((\mathbb{C}^\times, \times)\).
- Let \(n \in \ZZ\). Then \[ n\ZZ := \{ nz : z \in \ZZ\} \]is a subgroup of \((\ZZ,+)\).
- Let \(G=S_4\) and let \(V_4 = \{\id, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)\}\). You should check that \(V_4\) is a subgroup of \(S_4\) - it is called the Klein 4-group (or
*Viergruppe*in German, hence the notation). - Let \(A_n\) be the set of all even permutations in \(S_n\). Because the identity permutation is even, the product of two even permutations is even, and the inverse of an even permutation is even, \(A_n\) is a subgroup of \(S_n\), called the
**alternating group**. Furthermore \(|A_n|=n!/2\), since exactly half the permutations in \(S_n\) are even. - \(\{ e \}\) is a subgroup of \((G, *)\) for any group
*G*. This is called the**trivial subgroup**.*G*is a subgroup of \((G, *)\) for any group \((G, *)\). So every group has at least two subgroups.

**Definition 4.12**A subgroup of \((G,*)\) is called

**proper**if it is not equal to

*G*, and non-trivial if it is not equal to \(\{e \}\).

**Proposition 4.6**(The

**subgroup test**). Let \((G,*)\) be a group and let

*H*be a non-empty subset of

*G*such that for all \(x,y \in H\) we have \[\begin{equation} \tag{4.4} x^{-1}*y \in H. \end{equation}\] Then

*H*is a subgroup of

*G*.

*Proof. * Firstly *H* contains the identity element *e* of *G*. For since
*H* is non-empty there
is some element \(x\in H\), so by (4.4) with \(y=x\) we have \(x^{-1}*x \in H\), that is \(e \in H\).

Secondly, if \(x \in H\) then \(x^{-1} \in H\). For \(x,e \in H\) and so using (4.4) with \(y=e\) we get \(x^{-1}*e=x^{-1} \in H\).

Now if \(x,y \in H\) then \(x^{-1}\) is in*H*too, so \((x^{-1})^{-1} * y = x*y \in H\). Thus \(*\) gives a binary operation on

*H*. We have already seen that

*H*has the identity element, and each element of

*H*has its inverse in

*H*. Since \(*\) is associative on

*G*, it is certainly associative on

*H*. Thus

*H*is a group under \(*\), and so it is a subgroup of

*G*.

**Example 4.6 **If \(\sigma \in S_n\) is a permutation, we say \(\sigma\) **fixes** *i* if
\(\sigma(i)=i\). Let *H* be the subset of \(S_n\) consisting of all
permutations fixing *1*. We’ll show that *H* is a subgroup using the
subgroup test.

*H*is non-empty, since it certainly contains the identity permutation which fixes everything. Secondly, suppose \(x,y \in S_n\) fix 1. Then \(x(1)=1\), so \(x^{-1}(1) = 1\). Therefore \((x^{-1} \circ y) (1) = x^{-1}( y ( 1)) = x^{-1}(1) = 1\). It follows \(x^{-1}\circ y \in H\), and so

*H*is a subgroup of \(S_n\) by the subgroup test.

### 4.6.1 Cyclic subgroups

Let \((G,*)\) be a group, let \(g \in G\), and let \(\langle g \rangle = \{ g^i : i \in \ZZ\}\). Then \(\langle g \rangle\) is called the **cyclic subgroup generated by** *g*.

**Lemma 4.6**\(\langle g \rangle \leq G\).

*Proof.*Apply the subgroup test: \(\langle g \rangle\) contains

*g*so it’s certainly not empty, and if \(g^i, g^j \in \langle g \rangle\) then \((g^i)^{-1} * g^j = g^{-i}* g^j = g^{j-i} \in \langle g \rangle\). So the subgroup test implies \(\langle g \rangle \leq G\).

**Lemma 4.7**If

*g*has order

*n*, then \(\langle g \rangle\) has order

*n*.

*Proof. * We know from Lemma 4.3 that the elements \(e, g, g^2, \ldots, g^{n-1}\) of \(\langle g \rangle\) are all different,
so the size of \(\langle g \rangle\) is at least *n*. We’ll show
that it is exactly *n* by showing that any \(g^i \in \langle g \rangle\) equals one of these elements.

**Example 4.7 **

- Consider \(\langle (1,2) \rangle \leq S_3\). The permutation \((1,2)\) has order two (why?), so \(\langle (1,2)\rangle\) has size two, and its elements are \(\{ e, (1,2) \}\).
- The order of \([2]_6 \in \ZZ_6\) is 3, because \([2]+[2]=[4]\) and \([2]+[4]=[0]\), which is the identity element of \((\ZZ_6, +)\). Thus the cyclic subgroup generated by \([2]\) is \(\langle [2] \rangle = \{[0],[2],[4]\} \leq \ZZ_6\).