If a^(- 1) = a for every a in G then G is abelian.

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  So for any elements x,y in Cn, we have x=a^i, y=a^j for some integer i,j. Thus x*y=a^i*a^j=a^(i+j)=a^(j+i)=a^j*a^i=y*x. Thus Cn is abelian. 2. G is a group, then for every element a, it has an inverse a^(-1) in G.

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  is not divisible by |G| then H must contain a nontrivial subgroup of G." Proof: (a) If H is normal, then for any g in G, we have gHg^(-1)=H. Since |H|=2 and H=<a>={e,a}, then gag^(-1)=e or gag^(-1)=a. But if gag^(-1)=e, then a=e.

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