Question

Let a group g contain normal subgroups of order 3 and 5. Show that g has an element of order 15

Answer 1

Answer:

Let G be a group having unknown elements.

Let H and M be two normal subgroups of G having 3 and 5 elements.

H={p, q,e}, here $p^{-1}=q$, pq =e

M={p,q,r,s,e}, here  $p^{-1}=q$, $r^{-1}=s$,p q=e, r s=e

It means, subgroup M will contain element p, q,r,s,and apart from it, it will also contain pq, pr,ps, qr,qs,rs, pqr, qrs, prs, pqs and one element e.

As, M ⊂ G

M contains 5 elements following the property that , pq=pr=qr=qs=rs=sp=pqr=qrs=prs=pqs=e.

Therefore,these elements must be inside the set G.

G= {M, H, {p,q,r,s,pq, pr,ps, qr,qs,rs, pqr, qrs, prs, pqs , e}}.

Hence we can say that group G has one set containing  15 elements.