I'm currently making my way through "Groups and Symmetry", by M. A. Armstrong [ISBN 0387966757], and I'm stuck on what seems like it should be a very simple exercise. Page 14, problem 3.8, states:

However, I can't make this work unless I change the "or" to an "and". So, is this a misprint, or am I missing something blindingly obvious?

edit: never mind, maybe you're right. Anyway, there's a theorem that says the only cyclic multiplicative groups are the ones with prime order, which solves this problem. I think it's "and" because you need to have an even multiple of 11 to get 0 (mod22), which has no inverse and also isn't even in the group.

Well... Suppose that A is a subset of {1, 2, ..., 21} that contains at least one even number, and suppose that A forms a group (where the operation is multiplication modulo 22). Let 2k [tex]\in[/tex] A. 2k has to have an inverse, i.e a number x [tex]\in[/tex] A such that 2k * x = 1 (mod 22). But this is equivalent to saying that there exists a number y [tex]\in \mathbb{N}[/tex] such that 2kx - 1 = 22y. We see that 2 divides the right-hand side of the equation, but not the LHS, so we can never have equality (in the integers). Contradiction.

You can easily extend this argument to cover 11 as well.

This is basically a small special case of a more general theorem that says that only those numbers coprime to n, will have a multiplicative inverse modulo n (hope I phrased that correctly).

But does 1 have to be the identity? For the numbers x = 1, 11, and 12, we have x*x = x(mod22). We might be able to form a very small group with some number other than 1 as its identity. I'm sure there's a reason why that's wrong though.

lol maybe i should look these things up before posting. i was pretty sure i would make an ass of myself by responding in any way. maybe i was thinking of something to do with finite fields or something.

No it's not sufficient. I didn't check that 11 and 12 worked for any other numbers, but I was only provoking a proof that 1 must be the identity, otherwise the problem is rather simple. And I believe the set {1,11} works. 11 has two inverses, but I don't know if inverses have to be unique.

identities must be unique, inverses must be unique, you've demonstrated that the one element set {11} with mult mod 22 is the trivial group, as is {12}

in your example {1,11} there are two identities by your reasoning, that is not allowed, identities must be unique.

I was looking for something like this. At Mathworld.com, I don't see it explicitly stating that inverses and identites have to be unique, but I would naturally assume they are. At wikipedia.com, there is an article on group(mathematics) where some "Simple Theorems" are listed, including:

A group has exactly one identity element.

Every element has exactly one inverse.

But no proof for these. This seems like somthing that would arise more from definition than deduction, but if not, can you provide a proof or a link that has a proof?

AKG's proof: Suppose e and f are identity elements in a group G, then e=ef since f is an identity, and f=ef since e is an identity hence e=f, now the proof for the uniqueness of inverses is left as an easy exercise, and an important one.