# Thread: internal direct product question

1. ## internal direct product question

if m>2, must ZmXDn and Dmn always have different isomorphic types? (where X= internal direct product).
If it were the external direct product of the two, I can easily see that Zm+Dn is not isomorphic to Dmn, however i am not sure what to do with the internal direct product. I know that a group G is equal to the internal direct product of two groups if both of the two groups are normal in G and thei intersection is equal to e. Would I need this fact for this problem?
Any help would be great, I am really struggle with the concept of the internal direct product in this problem.
Thanks

2. Originally Posted by hopttoit
if m>2, must ZmXDn and Dmn always have different isomorphic types? (where X= internal direct product).
If it were the external direct product of the two, I can easily see that Zm+Dn is not isomorphic to Dmn, however i am not sure what to do with the internal direct product. I know that a group G is equal to the internal direct product of two groups if both of the two groups are normal in G and thei intersection is equal to e. Would I need this fact for this problem?
Any help would be great, I am really struggle with the concept of the internal direct product in this problem.
Thanks
I would suggest looking at the center of your group. $Z(\mathbb{Z}_m \times D_n) = \mathbb{Z}_m \times Z(D_n)$.

What does the center of $D_n$ and $D_{mn}$ look like? (It is different for even and off numbers).

3. the center of Dn when n is even is R0 and R180, when odd it is just Ro (identity element). So will ZmXDn and Dmn always have different centers? Iam thinking that Zm will always be in the center for ZmXDn and this would always make it different from Dmn.
Is this right?

4. Originally Posted by hopttoit
the center of Dn when n is even is R0 and R180, when odd it is just Ro (identity element). So will ZmXDn and Dmn always have different centers? Iam thinking that Zm will always be in the center for ZmXDn and this would always make it different from Dmn.
Is this right?
Yes, as I said earlier, $Z(\mathbb{Z}_m \times D_n) = \mathbb{Z}_m \times Z(D_n)$. However, you still need to do a tad more work - like if n is odd and m=2 then what?

Also, you have to take all this into the internal direct product...

5. So from definition of internal direct product, we know that Dmn= ZmXDn if Dmn is isomorphic to the external direct product of Zm and Dn (Zm+Dn). If n is odd and m equals 2, they would be isomorphic because we could identify two elements in Zm+Dn that satisfy the generator and relations rquirements for D2n.
I do not understand what you mean when you say that we need to take all this into the internal direct product....

6. Originally Posted by hopttoit
So from definition of internal direct product, we know that Dmn= ZmXDn if Dmn is isomorphic to the external direct product of Zm and Dn (Zm+Dn). If n is odd and m equals 2, they would be isomorphic because we could identify two elements in Zm+Dn that satisfy the generator and relations rquirements for D2n.
I do not understand what you mean when you say that we need to take all this into the internal direct product....
Well, you are doing this in the external direct product, you are looking at $\mathbb{Z}_m \times D_n$ not at a group $G$ with $H \cong \mathbb{Z}_m$, $K \cong D_n$ such that $H \lhd G$, $K \lhd G$ and $H \cap K = \{1\}$.

You have to show that such a group is not isomorphic to $D_{mn}$.