Separating groups into isomorphism classes.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

for #2. i know what all of them look like except for Z3 X Z2. How would i generate a group table for that one??

Shadow · **Posted:** Sat, 3 Mar 2012 08:30:19 UTC

DgrayMan wrote:

for #2. i know what all of them look like except for Z3 X Z2. How would i generate a group table for that one??

It's a product, you just list the pairs and add them in coordinates. I'm a bit confused though, none of those questions, in particular not #2, reference Z2 OR Z3...

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

Oops wrong pic!

Justin · **Joined:** Wed, 21 May 2003 04:27:18 UTC **Posts:** 995

DgrayMan wrote:

Oops wrong pic!

As Shadow said, nothing particularly difficult about $\mathbb{Z}_3 \times \mathbb{Z}_2$ ...it is the direct product of two prime order (cyclic) groups...that tells you a lot right there. For instance, the order of each element is the least common multiple of the orders of each "component" of the element. That implies, for instance, that this group has order 6.

The Cayley table is then easy to construct.

Shadow · **Posted:** Sat, 3 Mar 2012 22:40:43 UTC

Justin wrote:

DgrayMan wrote:

Oops wrong pic!

As Shadow said, nothing particularly difficult about $\mathbb{Z}_3 \times \mathbb{Z}_2$ ...it is the direct product of two prime order (cyclic) groups...that tells you a lot right there. For instance, the order of each element is the least common multiple of the orders of each "component" of the element. That implies, for instance, that this group has order 6.

The Cayley table is then easy to construct.

Since it's listed as a product, he doesn't need to worry about whether or not the original orders, and he can get order six just by multiplying the orders of the original group, it doesn't have to be cyclic for that to work.

To the op: You know the elements are (0,0), (0,1), (1,0), (1,1), (2,0), and (2,1), the table is easy, just add in components.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

so this is still using addition modulo 3 = {0 ,1 ,2} ? so (0,1)+(0,1)=(0,2) Or does it mean only 0 is mod3 and 1 is mod2? in (0,1), which in that case would make it (0,1)+(0,1)=(0,0)

Shadow · **Posted:** Sun, 4 Mar 2012 00:41:25 UTC

DgrayMan wrote:

so this is still using addition modulo 3 = {0 ,1 ,2} ? so (0,1)+(0,1)=(0,2) Or does it mean only 0 is mod3 and 1 is mod2? in (0,1), which in that case would make it (0,1)+(0,1)=(0,0)

The first component adds modulo 3 and the second adds modulo 2.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

Got it! phew!!!

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

By inspection, none of them look isomorphic? is this right??

Shadow · **Posted:** Sun, 4 Mar 2012 01:22:47 UTC

DgrayMan wrote:

By inspection, none of them look isomorphic? is this right??

No, that's not quite right. Now is the time to use Justin's advice and note that the element (1.1) in Z3 x Z2 maps to the element 1 in Z6 is an isomorphism.

Justin · **Joined:** Wed, 21 May 2003 04:27:18 UTC **Posts:** 995

Shadow wrote:

Since it's listed as a product, he doesn't need to worry about whether or not the original orders, and he can get order six just by multiplying the orders of the original group, it doesn't have to be cyclic for that to work.

True, but I think you can only multiply the orders when they are relatively prime...correct?

Shadow · **Posted:** Sun, 4 Mar 2012 01:28:33 UTC

Justin wrote:

Shadow wrote:

Since it's listed as a product, he doesn't need to worry about whether or not the original orders, and he can get order six just by multiplying the orders of the original group, it doesn't have to be cyclic for that to work.

True, but I think you can only multiply the orders when they are relatively prime...correct?

Nope, as a set $G_1\times G_2$ is just the cartesian product of G_1

and

, so the orders multiply no matter what they are. You are probably thinking of ways to combine them and generators when they're abelian to find new generators.

Justin · **Joined:** Wed, 21 May 2003 04:27:18 UTC **Posts:** 995

Shadow wrote:

Justin wrote:

Shadow wrote:

Since it's listed as a product, he doesn't need to worry about whether or not the original orders, and he can get order six just by multiplying the orders of the original group, it doesn't have to be cyclic for that to work.

True, but I think you can only multiply the orders when they are relatively prime...correct?

Nope, as a set $G_1\times G_2$ is just the cartesian product of G_1

and

, so the orders multiply no matter what they are. You are probably thinking of ways to combine them and generators when they're abelian to find new generators.

Ah, right...I guess I'm thinking of something related to the Fundamental Theorem of Abelian Groups(?)
The one where you characterize the (finite) Abelian groups as direct sums up to isomorphism...

Shadow · **Posted:** Sun, 4 Mar 2012 01:44:21 UTC

Justin wrote:

Shadow wrote:

Justin wrote:

Shadow wrote:

Since it's listed as a product, he doesn't need to worry about whether or not the original orders, and he can get order six just by multiplying the orders of the original group, it doesn't have to be cyclic for that to work.

True, but I think you can only multiply the orders when they are relatively prime...correct?

Nope, as a set $G_1\times G_2$ is just the cartesian product of G_1

and

, so the orders multiply no matter what they are. You are probably thinking of ways to combine them and generators when they're abelian to find new generators.

Ah, right...I guess I'm thinking of something related to the Fundamental Theorem of Abelian Groups(?)
The one where you characterize the (finite) Abelian groups as direct sums up to isomorphism...

Yes you are, and I suspected as much. Don't worry, this is a common mistake when thinking about these things before you really get them all the way down.

Justin · **Joined:** Wed, 21 May 2003 04:27:18 UTC **Posts:** 995

Shadow wrote:

Justin wrote:

Ah, right...I guess I'm thinking of something related to the Fundamental Theorem of Abelian Groups(?)
The one where you characterize the (finite) Abelian groups as direct sums up to isomorphism...

Yes you are, and I suspected as much. Don't worry, this is a common mistake when thinking about these things before you really get them all the way down.

LOL...I've been working on getting that for the past six years...some things in math are just harder to digest than others.

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Separating groups into isomorphism classes.

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