7.5.3(b)
H={0,3,6,9) => |H|=4.
|G|=12.
Therefore |G:H|=|G|/|H|=3.

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7.5.3(c)
Note that H={0,1,2,...,19}=G, since gcd(3,20)=1.
=> |G|=|H|=20.
Therefore |G:H|=|G|/|H|=1.

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7.5.9
Suppose H and K are nonequal subgroups of G of prime order p.
H^K is a subgroup of G since H and K are => since H^K is contained in H, it is a subgroup of H => |H^K| divides |H| => |H^K| is 1 or p. If it is p, then since |H| and |K| are also p, we get H=H^K=K, which contradicts to our assumption H!=K.
Therefore |H^K| must be 1 => H^K=.

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7.5.12
Suppose G is not cyclic. => it has no element of order 25. Let x be in G, x!=e. is a subgroup of G => || divides |G|=25.
But ||=|x|, x!=e, and we assumed G is not cyclic => |x|=5. So every non identity element of G has order 5. QED

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7.6.3(b)
Since |D₄|=8 and |N|=4, we get |D₄|/|N|=2 => there are 2 right and 2 left cosets of N in |D₄|. r0N={r0,r2,h,v}=Nr0, r1N={r1,r3,t,d}=Nr1. => N is normal.

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7.6.7
Let (g,h) be in GxH, and (a,e) be in G*. Then (g,h)(a,e)(g,h)^(-1)= (ga^(-1),heh^(-1))=(gag^(-1),e). This is in G* since g,a are in G and therefore so is gag^(-1). So, for any x in G*, y in GxH, we get yxy^(-1) is in G* => G* is normal.

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7.6.17
Since N and K are subgroups of G, so is N^K. But N^K is contained in K => N^K is a subgroup of K. Let k be in K, n be in N^K. Then since N is normal in G, knk^(-1) is in N. Also k and n are in K => knk^(-1) is in K => knk^(-1) is in N^K => N^K is normal in K.