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7.3.31
Let p,q be in K:=x ^-1Hx . => there are a,b in H s.t. p=x^-1ax, q=x^^-1bx. Then pq=x^-1abx. a,b are in H, and H<G => ab is in H => x^-1abx is in K. I.e. pq is in K. p^-1=(x^-1ax)^-1 =x^-1a^-1x. a is in H, H<G => a^-1 is in H => x^-1a^-1x is in K, i.e. p^-1 is in K. So p,q in K => pq and p^-1 are in K => K < G.

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7.4.5
U₅={1,2,3,4} and U₁₀={1,3,7,9}. Note that the element 2 is a generator for U₅ and 3 is a generator for U₁₀ => U₅ and U₁₀ are cyclic. Since both U₅ and U₁₀ have order 4, they are isomorphic to Z₄ => they are isomorphic to each other.

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7.4.9
f,g are isomorphisms => they are bijections => gf is a bijection. Let a,b be in G. Then gf(ab)=g(f(ab))=g(f(a)f(b))=g(f(a))g(f(b))=gf(a)gf(b) => gf is an isomorphism.

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It is easy to prove the following:
Lemma: If f:H->G is an isomorphism => for any a in H |a|=|f(a)|.
Corollary: The number of elements of given order n in isomorphic groups G and H is the same.
Proof of Lemma:
>
Let n=|a|, m=|f(a)|. Then e_G=f(e_H)= f(aⁿ)=f(a)ⁿ => m|n. f(a^m)=f(a)^m= e_G. f is an isomorphism => a^m=e_H => n|m. n|m, m|n => n=m.
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WARNING: Some people used this or similar results in their homework and this time I didn't take points off, but in the future you should prove this kind of results if you want to use them.

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The following 3 problems can be done using this result.

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7.4.28a
Z₆ has an element of order 6, namely 1, but S₃ doesn't.

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7.4.28c
(1,0) is of order 4 in Z₄xZ₂ but Z₂xZ₂xZ₂ has no element of order 4.

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7.4.31
D₄ and the quaternions have a different number of elements of order 4.