# Math Help - A series of questions...(1)

1. ## A series of questions...(1)

Let G,H,K be finitely generated abelian groups. Show that if G*H is isomorphic to G*K , then H is isomorphic to K, where * represents the external direct product of groups.
One simple idea of this problem is to divide G,H,K into Z(p1^k1)*Z(p2^k2)...,but the question is to prove the unity of such representation, that is to say, is it possible that Z(6)*Z(2) isomorphic to Z(2)*Z(2)*Z(3), or the more general situation?
Thank you!

2. Originally Posted by ynj
Let G,H,K be finitely generated abelian groups. Show that if G*H is isomorphic to G*K , then H is isomorphic to K, where * represents the external direct product of groups.
One simple idea of this problem is to divide G,H,K into Z(p1^k1)*Z(p2^k2)...,but the question is to prove the unity of such representation, that is to say, is it possible that Z(6)*Z(2) isomorphic to Z(2)*Z(2)*Z(3), or the more general situation?
Thank you!
Can;t you just factor out G, and then use the first isomorphism theorem?

3. ?Can you give more ditail?thank you

4. Originally Posted by ynj
Let G,H,K be finitely generated abelian groups. Show that if G*H is isomorphic to G*K , then H is isomorphic to K, where * represents the external direct product of groups.
One simple idea of this problem is to divide G,H,K into Z(p1^k1)*Z(p2^k2)...,but the question is to prove the unity of such representation, that is to say, is it possible that Z(6)*Z(2) isomorphic to Z(2)*Z(2)*Z(3), or the more general situation?
Thank you!

This is actually true. $\mathbb{Z}_{nm}=\mathbb{Z}_n \times \mathbb{Z}_m \Leftrightarrow gcd(m,n)=1$

As far as proving this I don't think it is that complicated.

If $G\times H \cong G \times K$, then there is an isomorphism $\Phi: G \times H \rightarrow G \times K$ so say
$\Phi(g,h)=(g',k)$, but then just look at the projection onto the second coordinate of this map. It inherits surjectivity, homomorphism, welldefinedness, from the fact that $\Phi$ is an isomorphism.
It will go from $\pi_2\circ \Phi:G\times H \rightarrow \{0\}\times K \cong K$, it clearly has kernel G, and $G\times H/G\cong H$

5. Originally Posted by ynj
?Can you give more ditail?thank you
Well, note that $(G \times H)/G \cong H$.

I don't understand where finitely generated or abelian comes in though...?

6. Originally Posted by Gamma
it clearly has kernel G
why? do you mean (g,e)?since G is not a subset of G*H...
we may write the isomorphism as F(g,h)=(f1(g,h),f2(g,h))
and we have f1(g1g2,h1h2)=f1(g1,h1)f1(g2,h2)
f2(g1g2,h1h2)=f2(g1,h1)f2(g2,h2)
how to ge the kernel? that is, to solve the equaltion f2(g,h)=e??
do you mean f2(g,e)=e?but what i can deduce is f2(e,e)=e??
I am confused..

7. yeah, kernel needs to be $G\times e$, you are right, but this is pretty obviously isomorphic to just G by just dropping the e part. It can be confusing sometimes when you are dealing with something so obviously true I think. If you believe you can just write the isomorphism in terms of the coordinate isomorphism, I would just take the second coordinate function and show why that must be bijective and a homomorphism, that might be easier than dealing with first isomorphism theorem stuff.

I was not sure you were allowed to assume that or not, so I went that route instead.

8. To be honest I am not even sure this is necessary, I think you are making it way too hard. If $G\times H \cong G\times K$, then $\frac{G\times H}{G\times\{1\}} \cong \frac{G\times K}{G\times \{1\}}$
Thus $H\cong \frac{G\times H}{G\times \{1\}} \cong \frac{G\times K}{G\times \{1\}} \cong K$

9. Originally Posted by Gamma
To be honest I am not even sure this is necessary, I think you are making it way too hard. If $G\times H \cong G\times K$, then $\frac{G\times H}{G\times\{1\}} \cong \frac{G\times K}{G\times \{1\}}$
Thus $H\cong \frac{G\times H}{G\times \{1\}} \cong \frac{G\times K}{G\times \{1\}} \cong K$
maybe I am wrong, you can first prove that $\frac{G\times H}{G\times\{1\}} \cong \frac{G\times K}{G\times \{1\}}$, the isomorphism is well-defined.
I think it is not so easy to prove.
if (g1,h1)(G,e)=(g2,h2)(G,e), then (g1g2^(-1),h1h2^(-1))is in (G,e). Thus h1=h2,g1g2^(-1) is in G. Note that ,we cannot say g1=g2, right?
then we shall prove (f1(g1,h1),f2(g1,h1))(G,e)=(f1(g2,h2),f2(g2,h2))(G ,e)
this will require f2(g1g2^(-1),e)=e. do you mean f2(g,e)=e???why?

10. if we remove the condition "finitely generated", the claim in the problem will no longer be true. for example let $G$ be the direct product of infinitely many copies of $\mathbb{Z}$ and let $H=\mathbb{Z}$ and

$K=\mathbb{Z} \times \mathbb{Z}.$ then $G \times H \cong G \times K$ but $H \ncong K.$

the problem can be reduced to "finite" abelian groups. the reason is that we have $G=\mathbb{Z}^r \times G_1, \ H=\mathbb{Z}^s \times H_1, \ K=\mathbb{Z}^t \times K_1,$ where $r,s,t \geq 0$ are some integers and $G_1,H_1,K_1$ are

some finite abelian groups. then $G \times H \cong G \times K$ gives us $s=t$ and $G_1 \times H_1 \cong G_1 \times K_1.$ see if you can solve the problem now!

11. Originally Posted by NonCommAlg
if we remove the condition "finitely generated", the claim in the problem will no longer be true. for example let $G$ be the direct product of infinitely many copies of $\mathbb{Z}$ and let $H=\mathbb{Z}$ and

$K=\mathbb{Z} \times \mathbb{Z}.$ then $G \times H \cong G \times K$ but $H \ncong K.$

the problem can be reduced to "finite" abelian groups. the reason is that we have $G=\mathbb{Z}^r \times G_1, \ H=\mathbb{Z}^s \times H_1, \ K=\mathbb{Z}^t \times K_1,$ where $r,s,t \geq 0$ are some integers and $G_1,H_1,K_1$ are

some finite abelian groups. then $G \times H \cong G \times K$ gives us $s=t$ and $G_1 \times H_1 \cong G_1 \times K_1.$ see if you can solve the problem now!
yeah,I know this result...my original question is to ask whether is it possible a finite abelian group has two way of decomposition..like Z(18)=Z(9)*Z(2)?=Z(3)*Z(3)*Z(2)..though this example is impossible, how to prove in general?

12. Originally Posted by ynj
yeah,I know this result...my original question is to ask whether is it possible a finite abelian group has two way of decomposition..like Z(18)=Z(9)*Z(2)?=Z(3)*Z(3)*Z(2)..though this example is impossible, how to prove in general?
Do you mean something like $C_6 \cong C_2 \times C_3$? This holds because 2 and 3 are coprime...

Thank you very much for your (sketch) proof NonCommAlg - one day I will be less ignorant of the infinite...

13. Originally Posted by Swlabr
Do you mean something like $C_6 \cong C_2 \times C_3$? This holds because 2 and 3 are coprime...

Thank you very much for your (sketch) proof NonCommAlg - one day I will be less ignorant of the infinite...
not really. I mean if n=p1^k1*p2^k2....pm^km=q1^r1*q2^r2....qm^rm,where pi,qi are primes(not all the same),is it possible that Z(n)=Z(p1^k1)....Z(pm^km)=Z(q1^r1)...Z(qm^rm)?

14. Originally Posted by ynj
not really. I mean if n=p1^k1*p2^k2....pm^km=q1^r1*q2^r2....qm^rm,where pi,qi are primes(not all the same),is it possible that Z(n)=Z(p1^k1)....Z(pm^km)=Z(q1^r1)...Z(qm^rm)?
well-done! you have managed to confuse everybody in here! haha ... what do you mean by "not all"? are you saying that we might have $p_1=p_2$ for example? in this case the answer is no!

anyway, to solve your problem, you should follow the idea that i gave you in my previous post. as i said, we may assume that $G,H,K$ are finite abelian. now since every finite abelian group is a

product of cyclic subgroups, we'll get another reduction: we may assume that $G$ is cyclic.

15. I am going to take a shot at trying to understand what you are asking.

Are you asking us whether or not the representation in terms of products of cyclic groups is unique?

In general the answer is no as we have given several examples

$\mathbb{Z}_{mn} \cong \mathbb{Z}_m \times \mathbb{n}$ if and only if $gcd(m,n)=1$.

To rectify this question of multiple representations of the same group, we use invariant factor or elementary divisor decomposition, along with some division requirements.

For any finitely generated abelian group, we have
$G\cong \mathbb{Z}^r \times \mathbb{Z}_{k_1} \times \mathbb{Z}_{k_2} \times \cdots \times \mathbb{Z}_{k_n}$
where $k_1|K_2|...k_{n-1}|k_n$.

Then the free rank r and the invariant factors $k_1,k_2,...,k_n$ are uniquely determined.

Similarly, there is a uniqueness to the elementary divisor decomposition too, and it comes from essentially decomposing the elementary divisors using the isomorphism I first mentioned above into powers of primes. Then this list too is unique, up to rearranging the order of the primes.

$G\cong \mathbb{Z}^r \times \mathbb{Z}_{p_1^{k_1}} \times \mathbb{Z}_{p_2^{k_2}} \times \cdots \times \mathbb{Z}_{p_n^{k_n}}$ where $p_i^{k_i}$ is a power of a prime $p_i$ where the $p_i$ are not necessarily distince primes.

You could make this unique too, by saying list the largest primes first, within those primes list the highest exponent first and then go on down until you have exhausted the list.

This is if two groups are isomorphic, then then must have the same free rank and the same invariant factor decomposition and elementary divisor decomposition (this one up to reordering).

If they do not have the same free rank, or have different invariant factor decomposition or have different elementry divisor decompositions, then they are not isomorphic.

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