# Subgroups

**Subgroups and subgroup tests**: A subgroup of a group G is a subset of G which is a subgroup in its own right (with the same group operation).

There are two subgroup tests, resembling the two subring tests.

**Proposition 3.9 **(First Subgroup Test) A non-empty subset H of a group G is a subgroup of G if, for any h,k ∈ H, we have hk ∈ H and h^{−1} ∈ H.

**Proof **We have to show that H satisfies the group axioms. The conditions of the test show that it is closed under composition (G0) and inverses (G3). The associative law (G1) holds in H because it holds for all elements of G.

We have only to prove (G2), the identity axiom. We are given that H is non-empty, so choose h ∈ H. Then by assumption, h^{−1} ∈ H, and then (choosing k = h^{−1}) 1 = hh^{−1} ∈ H. We can reduce the number of things to be checked from two to one:

**Proposition 3.10** (Second Subgroup Test) A non-empty subset H of a group G is a subgroup of G if, for any h,k ∈ H, we have hk^{−1} ∈ H.

**Proof** Choosing k = h, we see that 1 = hh^{−1} ∈ H. Now using 1 and h in place of h and k, we see that h^{−1} = 1h^{−1} ∈ H. Finally, given h,k ∈ H, we know that k^{−1} ∈ H, so hk = h(k^{−1})^{−1} ∈ H. So the conditions of the First Subgroup Test hold.

**Example** Look back to the Cayley tables in the last chapter. In the first case, {e,y} is a subgroup. In the second case, {e,a}, {e,b} and {e,c} are all subgroups.

**Cyclic groups**: If g is an element of a group G, we define the powers gn of G (for n∈ Z) as follows: if n is positive, then gn is the product of n factors g; g0 = 1; and g^{−n} = (g^{−1})n.

The usual laws of exponents hold: g^{m n} = g^{m} · g^{n} and g^{mn} = (g^{m})^{n}.

A cyclic group is a group C which consists of all the powers (positive and negative) of a single element. If C consists of all the powers of g, then we write C = (g), and say that C is generated by g.

**Proposition 3.11 **A cyclic group is Abelian.

**Proof** Let C = hgi. Take two elements of C, say g^{m} and g^{n}. Then

g^{m} · g^{n} = g^{m n} = g^{n} · g^{m}.

Let C = (g). Recall the order of g, the smallest positive integer n such that

g^{n} = 1 (if such n exists – otherwise the order is infinite).

**Proposition 3.12 **Let g be an element of the a group G. Then the set of all powers (positive and negative) of g forms a cyclic subgroup of G. Its order is equal to the order of g.

**Proof **Let C = {g^{n} : n ∈ Z}. We apply the Second Subgroup test: if gm,gn 2 C, then (g^{m})(g^{n})^{−1} = g^{m−n} ∈ C. So C is a subgroup. If g has infinite order, then no positive power of g is equal to 1. It follows that all the powers gn for n 2 Z are different elements. (For if g^{m} = g^{n}, with m > n, then g^{n−m} = 1.) So C is infinite.

Suppose that g has finite order n. We claim that any power of g is equal to one of the elements g^{0} = 1,g^{1} = g, . . . ,g^{n−1}. Take any power gm. Using the division algorithm in Z, write m = n^{q r}, where 0 r n^{−1}. Then

g^{m} = g^{nq r} = (g^{n})^{q} · g^{r} = 1 · g^{r} = g^{r}.

Furthermore, the elements 1,g, . . . ,g^{n−1} are all different; for if g^{r} = g^{s}, with 0 r < s n−1, then g^{s−r} = 1, and 0 < s−r < n, contradicting the fact that n is the order of g (the smallest exponent i such that g^{i} = 1).

**Cosets**: Given any subgroup H of a group G, we can construct a partition of G into “cosets” of H, just as we did for rings. But for groups, things are a bit more complicated.