Isomorphism type of a finite group with respect to multiplication modulo 65

Question

I'm the same guy revising for my group theory exam and posted a few days ago. I'm at the chapter on Finitely Generated Abelian Groups, and my prof gave this example which I don't quite understand:

Let $G = \{1,8,12,14,18,21,27,31,34,38,44,47,51,53,57,64 \} $ and the group operator be multiplication modulo 65. What is G isomorphic to?

This example was given at the end of chapter and I think he was trying to use it to tie in the corollaries of the Fundamental Theorem of Finitely Generated Abelian Groups. Here's his solution:

"Observe that $G$ is an abelian group of order 16. Hence, the possibilities (up to isomorphism) are:

• $\mathbb{Z}_{16}$ (False, since it has no element of order 16)

• $\mathbb{Z}_{2} \times \mathbb{Z}_{8}$ (False, since $(0,1)$ has order 8)

• $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{4}$ (False, since $(1,0,0), (0,1,0),(1,1,0),(0,0,2)$ have order 2)

• $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$ (False, since all elements except $(0,0,0,0)$ have order 2)

$\therefore G \cong \mathbb{Z}_{4} \times \mathbb{Z}_{4}$."

What I think I understand is that since $G$ has 16 elements, the set $S$ that generates it, i.e. $<S>$, needs to have at least 1 element that has order 16. (right?) If I'm right, then I understand why the last two points are false. Isn't $1 \in \mathbb{Z}_{16}$ and element that has order 16? Also, why does $\mathbb{Z}_{2} \times \mathbb{Z}_{8}$ need only 1 element for counterexample?

Answer 1

If $G$ has order $n$, then it need not have an element of order $n$. The Klein 4-group supplies an example of this (it is a group of order 4 in which every nonidentity element has order 2). A positive result along these lines is Cauchy's Theorem, which states that $G$ will have an element of order $p$ if $p$ is a prime dividing the order of $G$.

As for your last question, if $G$ were isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_8$, then the image of $(0,1)$ in $G$ under such an isomorphism would have order 8. Show that no element of $G$ has order 8. Then no such isomorphism can exist.

Answer 2

In order to appreciate the reasons given in the enumeration (which all boil down to saying that the suggested abelian group has a certain number of elements of certain order), you should compute the order of all elelments of $G$. For example, the powers of $8$ are $8^2=64$, then $8^3=57$, then $8^4=1$, so the order of $8$ is $4$. Once you have done this for all elements, you will indeed verify that $G$ has no element of order $16$, no element of order $8$, no three elements of order $2$, etc.