# Isomorphic Type

1. Sep 5, 2010

### roam

1. The problem statement, all variables and given/known data

[PLAIN]http://img541.imageshack.us/img541/9880/34132542.gif [Broken]

3. The attempt at a solution

For part (a), I think since the order of an element g is the smallest integer n such that gn=e, we will have:

8n mod 65 = 1 => n=4

64n mod 65 = 1 => n=2

14n mod 65 =1 => n=2

Am I right so far?

Now for part (b), what is it meant by an "isomorphic type"? And what do I need to do?

Last edited by a moderator: May 4, 2017
2. Sep 5, 2010

### HallsofIvy

Staff Emeritus
It means, I believe the "type" of group that G is isomorphic to. Since G has 8 elements, it is, of course, isomorphic to a group of order $8= 2^3$. What can you say about groups with exactly 8 elements. How many different "kinds" are there?

3. Sep 5, 2010

### roam

Do you mean the following direct products:

$$\mathbb{Z}_{2^3}$$

$$\mathbb{Z}_{2^2} \oplus \mathbb{Z}_{2}$$

$$\mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{2}$$

Is that right?

4. Sep 6, 2010

### HallsofIvy

Staff Emeritus
Yes. Now which of those is your G isomorphic to?

5. Sep 7, 2010

### roam

G is a subgroup of U(65), not all of U(65). So I think

$$U(65) \cong U(5.13) \cong U(5) \oplus U(13) \cong \mathbb{Z}_4 \oplus \mathbb{Z}_{12}$$

But the problem is that none of the 3 groups in my previous post will then be isomorphic to $$\mathbb{Z}_4 \oplus \mathbb{Z}_{12}$$. E.g. since lcm(12,4)=12, $$\mathbb{Z}_4 \oplus \mathbb{Z}_{12}$$ has elements of order 12, but $$\mathbb{Z}_{2^3}$$ has no elements of order 12. So they are not isomorphic.

So what is wrong? I'm confused...

6. Sep 26, 2010

### roam

HallsofIvy, "by finding the isomorphic type" they mean write G as an external product of cyclic groups of prime power order.

So, G is an abelian group of order 8=23. The fundamental theorem of Finite Abelian Groups implies that either $$G \cong \mathbb{Z}_8$$ or $$\mathbb{Z}_4 \oplus \mathbb{Z}_2$$ or $$\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$$.

From my work in part (a) I know that G has 2 distinct subgroups of order 2: $$\left\langle14 \right\rangle = \{ 1,14 \}$$ and $$\left\langle 64 \right\rangle= \{ 1,64 \}$$, and one subgroup of order 4: $$\left\langle 8 \right\rangle = \{ 1,8,57,64 \}$$.

Does it follow that G is non-cyclic and $$G \cong \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_4$$?