# Groups of prime order structurally distinct?

1. Oct 16, 2012

### dumbQuestion

I have a question. If I have a group G of order p where p is prime, I know from the *fundamental theorem of finite abelian groups* that G is isomorphic to Zp (since p is the unique prime factorization of p, and I know this because G is finite order) also I know G is isomorphic to Cp (the pth roots of unity). I also know that G is cyclic and since its isomorphic to Zp I know that all of its elements are generators. Also we know that the only subgroups of G are the trivial subgroup and G itself. We know because G is cyclic, that it is abelian. These are the properties I can determine. [EDIT: Thought of another thing. any nontrivial homomorphism from h: G --> G' should be injective right because ker(h) should be trivial because ker(h) is a subgroup of G and we know G only has subgroups {e} and G itself, and if h is not trivial this means ker(h) must be {e} so its trivial meaning h is injective]

But is it true that for each prime p, there is only one structurally distinct group (up to isomorphism)? Is there a theorem that indicates one way or another this fact?

EDIT: Think I figured it out. FUndamental theorem of finite abelian groups gauranteeds that if G and G' are both groups of order p where p is prime, then G isomorphic to Zp and G' isomorphic to Zp so this means G isomorphic to G'. Since the selection of G and G' are arbitrary, this means for all G, G' of order p, p prime that G is isomorphic to G' so there is only one structurally distinct group of order p

Last edited: Oct 16, 2012
2. Oct 16, 2012

### jgens

If G is a group of order p, then define f:G → Zp by mapping a generator of G to 1. This produces an isomorphism.

3. Oct 16, 2012

### dumbQuestion

But this is showing that G and Zp are isomorphic, right? I'm curious about two groups of order p that are not isomorphic to each other.

4. Oct 16, 2012

### Number Nine

Isomorphism is an equivalence relation. If both groups of order p are isomorphic to Zp, then they are isomorphic to each other.

5. Oct 16, 2012

### dumbQuestion

Yeah I see that now, I feel kind of stupid now for not seeing it before!