**Problem**. Let be a group, a prime number and Prove that either is cyclic or where is the dihedral group of order

**Solution**. It is clear for So we will assume that is an odd prime. Choose with Let Since every subgroup of index 2 is normal, is a normal subgroup of and thus for some integer Note that, since is odd, and hence Now

because Thus and hence because Therefore either or So we will consider two cases:

*Case 1* : In this case and so Thus is abelian and hence because is odd. So is cyclic in this case.

*Case 2* : In this case and so

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Hi, thanks for sharing your proof 🙂 I just don’t understand how you can conclude that G= HK. Is there a crucial Theorem I miss ? Doesn’t look obvious to me

Hi, thanks so much for this proof. In case 2, we say that G has the given presentation but how do we know that there are no other relations which a and b must satisfy?

Let be the generators of which satisfy the relations . Since satisfy the same realtions, there exists a group homomorphism sending to respectively. Obviously is surjective because generate Thus . But and so i.e. is injective as well.