## 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 $$h^{-1} \in H$$.
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 $$$\tag{4.4} x^{-1}*y \in H.$$$ 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.

### 4.6.1 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.

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