# Quiz 1 solutions 1. (a) What is a group?

```Quiz 1 solutions
1. (a) What is a group?
A group is a set G with an operation &middot; that satisfies
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Closure: a, b ∈ G ⇒ a &middot; b ∈ G
Associativity: a, b, c ∈ G ⇒ (a &middot; b) &middot; c = a &middot; (b &middot; c) ∈ G
Identity: there is an e ∈ G such that a &middot; e = e &middot; a = a for all a ∈ G
Inverses: for every a ∈ G, there is an a−1 ∈ G such that a &middot; a−1 = a−1 &middot; a = e.
(b) Suppose N is a subgroup of a group G. When is N normal? (Give just one of the equivalent
conditions we discussed in class.)
One of:
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gN g −1 ⊂ N for all g ∈ G
gN g −1 = N for all g ∈ G
every left coset is a right coset
the product of any two right cosets is a right coset
2. Let G = (Q, +) be the group of rational numbers, with addition as its operation. Let H = (Z, +) be the
subgroup of integers. Show that H has infinite index in G.
For any rational number q, Z + q is a coset of H in G, and Z + q = Z + r if and only if q − r is an
integer. Therefore, Z + 1/2, Z + 1/3, Z + 1/4, . . . are all distinct cosets of H. There are infinitely many
of these, so [G : H] = ∞.
3. Suppose that G is a group of order 60, H is a subgroup of order 5, and K is a subgroup of order 12.
Show that G = HK.
We want to show that o(HK) = 60. We have
o(HK) =
60
o(H)o(K)
=
o(H ∩ K)
o(H ∩ K)
so we need to show that H ∩ K = {e}. It is a subgroup of H, so o(H ∩ K) is a divisor of 5, and it
is a subgroup of K, so o(H ∩ K) is a divisor of 12. The only common divisor of 5 and 12 is 1, so
o(H ∩ K) = 1.
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