4.5 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 \(h^{-1} \in H\).

Then we say that 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.

Firstly, 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.

4.5.1 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 4.7 \(\langle g \rangle \leq G\).

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 4.8 If g has order n, then \(\langle g \rangle\) has order n.

Proof. We know from Lemma 4.4 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.

We can write \(i = qn + r\) where \(q,r \in \ZZ\) and \(0 \leq r < n\) by the division algorithm. So \[ g^i = g^{qn+r} = g^{qn}*g^r = (g^n)^q *g^r = e^q * g^r = e*g^r = g^r.\] Thus any element of \(\langle g \rangle\) equals one of \(e, g, \ldots, g^{n-1}\), and we’re done.

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 \(\langle [2] \rangle = \{[0],[2],[4]\} \leq \ZZ_6\).