7. (1 mark) If G is the additive group of integers and H is a subgroup of G, defined
by H = {5x|x E G}, find the distinct left cosets of H in G.
Answers
Step-by-step explanation:
Lemma. Let G,H be finite groups. Then gcd(|G|,|H|)=1 if and only if any subgroup of G×H is of the form G′×H′ for some subgroups G′⊆G and H′⊆H.
Proof. "⇒" Let K be a subgroup of G×H. Let eG,eH be neutral elements of G,H respectively. Now let (x,y)∈K. Since gcd(|G|,|H|)=1 then it follows that the order of y and y|G| are the same. Thus y|G| generates ⟨y⟩. In particular we have
(x,y)∈K ⇒
(x|G|,y|G|)=(eG,y|G|)∈K ⇒
(eG,y)∈K
The last implication because y can be generated from y|G|. Analogously if (x,y)∈K then (x,eY)∈K.
Now if you consider projections
πG:G×H→G
πH:G×H→H
then you (obviously) always have
K⊆πG(K)×πH(K)
What we've shown is the oppositie inclusion which completes the proof.
"⇐". Assume that d=gcd(|G|,|H|)≠1. Let p|d be a prime divisor and pick elements x∈G,y∈H such that |x|=|y|=p (they exist by the Cauchy's theorem). Now consider a subgroup K of G×H generated by (x,y). This subgroup is of prime order p. In particular if K=G′×H′ then either G′ or H′ has to be trivial (because the order of product is equal to product of orders and p is prime). But that is impossible since (x,y)∈K and none of x,y is trivial. Contradiction. □
So what it means is that if the orders of G and H are not relatively prime then there is a subgroup of G×H that is not a product of subgroups of G and H. This should give you plenty of examples, e.g.
G=H=Zn
K=⟨(1,1)⟩
Answer:
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