Exploring Order of Elements in Groups: Part (a) & (b)

[roam]

Homework Statement

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

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?

 

[HallsofIvy]

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?

 

[roam]

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?

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?

 

[HallsofIvy]

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?

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

 

[roam]

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

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...

 

[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 $$G \cong \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_4$$?