7. COSETS AND LAGRANGE’S THEOREM
99
|H||K|
). For two finite subgroups H and
|H \ K|
K of a group G, define the set HK = {hk | h 2 H, k 2 K}. Then
|H||K|
|HK| =
.
|H \ K|
Theorem (7.2 — |HK| =
Proof.
Although the set HK has |H||K| products, all not need be distinct. That is,
we may have hk = h0k 0 where h 6= h0 and k 6= k 0. To determine |HK|, we
need to determine the extent to which this happens. For every t 2 H \ K,
hk = (ht)(t 1k), so each group element in HK is represented by at least
|H \ K| products in HK. But hk = h0k 0 =) t = h 1h0 = kk 0 1 2 H \ K,
so h0 = ht and k 0 = t 1k. Thus each element in HK is represented by exactly
|H||K|
|H \ K| products, and so |HK| =
.
⇤
|H \ K|
Problem (Page 158 # 41). Let G be a group of order 100 that has a
subgroup H of order 25. Prove that every element of order 5 in G is in H.
Solution.
Let a 2 G and |a| = 5. Then by Theorem 7.2 the set haiH has exactly
5 · |H|
elements and |hai \ H| divides |hai| = 5 =) |hai \ H| = 5 =)
|hai \ H|
hai \ H = hai =) a 2 H.
⇤