1. ## Isomorphic

Let G be any group with no proper , nontrivial subgroups and assume abs value(G)>1. Prove that G must be isomorphic to Z_p for some prime p.

I know we have an isomorphism if a group is 1-1, onto, and the homomorphism property holds.

G is cyclic means G=<a>, where a is a generator.

I know a cyclic group of order n has exactly one subgroup of order m for each positive divisor m of n.

2. Write $\displaystyle |G|=n>1$. Let $\displaystyle a\ne e$. What can be said about the order of $\displaystyle \langle a \rangle$?

3. has exactly one subgroup of order m for each positive divisor m of n

4. First, you don't know that $\displaystyle G$ is cyclic. Second, you didn't answer the question. What is the order/size of $\displaystyle \langle a\rangle$?

5. is it infinite order since there is no n such that a^n=e

6. I took that your group $\displaystyle G$ was finite. If this is so, then $\displaystyle \langle a\rangle$ cannot be infinite!

7. then it has finite order

8. You should be able to pin it down, though. In particular, can $\displaystyle 1<\langle a\rangle<|G|$?

9. I guess i"m confused on how to pin it down.

10. order is n

11. Precisely. If the order of $\displaystyle \langle a\rangle$ were in between 1 and $\displaystyle n$, it would be a proper subgroup, contrary to your assumption. This means that $\displaystyle \langle a\rangle=G$; in other words, $\displaystyle G$ is cyclic. Now you just need to show that $\displaystyle |G|$ is prime.