*A Book of Abstract Algebra*by Charles Pinter. I ran into a snag on page 54. This is problem C5:

"Let G be a

*finite*group and let S be a nonempty subset of G. Suppose S is closed with respect to multiplications. Prove that S is a subgroup of . (Hint: it remains to prove that S contains e and is closed with respect to inverses. Let \(\displaystyle S = \left\{ {{a_1}, \ldots {a_n}} \right\}\). If \(\displaystyle {a_i} \in S\), consider the

*distinct*elements \(\displaystyle {a_i}{a_1},{a_i}{a_2}, \ldots {a_i}{a_n}\).)"

OK ... I did figure out part of it. If I imagine a Cayley Table for S, the set of elements \(\displaystyle {a_i}{a_1},{a_i}{a_2}, \ldots {a_i}{a_n}\) will be a row in it. That row will be a selection from the corresponding row of the Cayley Table for G. No row in the Cayley Table for G has any duplicates, so no selection from it can have any duplicates either. If S is closed under multiplication, then \(\displaystyle {a_i} \in S\), then \(\displaystyle {a_i}\) must appear in that row somewhere. For some k, \(\displaystyle {a_i} = {a_i}{a_k}\) and then \(\displaystyle e ={a_k}\).

Or does it? Actually, I don't think so. I think that last step fails.

Here's a sample Cayley Table, not a group, but a binary structure on the set \(\displaystyle \left\{ {a,b,c} \right\}\):

Every element appears in every row and column, nothing appears in any cell that isn't an element of the set, but there's no identity.

Either I disproved what Dr. PInter wants me to prove, or I made a mistake. Can anybody show me where I went wrong?

Thanks in advance.