Problem about a group with two inner direct product representations

[stragequark]

The problem: Suppose G is Abelian with two representations as the internal direct product of subgroups: G=HxK1, G=HxK2. Assume K1 is a subset of K2 and show K1=K2.
My attempted solution: I took the element (e_H, k_2), where e_H is the identity element of H and k_2 is an arbitrary element in K2, and multiplied it by an arbitrary element of HxK1 which I called (h,k_1) and I said the product (h, k_1*k_2) is an element of G=HxK1 and hence k_1*k_2 is an element of K1. Since K1 is closed under inverses I know that k_2 is an element of K1 since I can multiply k_1*k_2 by k_1 inverse. so K2 is a subset of K1.

Alternatively I thought that the order of G would be ord(H)ord(K1) and also ord(H)ord(K2), so ord(K1)=ord(K2). I am not given that these are finite, but may it still follow that if ord(K1)=ord(K2) and K1 is a subset of K2 that K1=K2?

I am suspicious of my proof because it doesn't use some of the problem's assumptions.
Your help is much appreciated.

 

[Euge]

The problem: Suppose G is Abelian with two representations as the internal direct product of subgroups: G=HxK1, G=HxK2. Assume K1 is a subset of K2 and show K1=K2.
My attempted solution: I took the element (e_H, k_2), where e_H is the identity element of H and k_2 is an arbitrary element in K2, and multiplied it by an arbitrary element of HxK1 which I called (h,k_1) and I said the product (h, k_1*k_2) is an element of G=HxK1 and hence k_1*k_2 is an element of K1. Since K1 is closed under inverses I know that k_2 is an element of K1 since I can multiply k_1*k_2 by k_1 inverse. so K2 is a subset of K1.

Alternatively I thought that the order of G would be ord(H)ord(K1) and also ord(H)ord(K2), so ord(K1)=ord(K2). I am not given that these are finite, but may it still follow that if ord(K1)=ord(K2) and K1 is a subset of K2 that K1=K2?

I am suspicious of my proof because it doesn't use some of the problem's assumptions.
Your help is much appreciated.

Hi stragequark,

Your attempted proof is actually correct (you can just say at the end that since $K_2 \subset K_1$ as well, $K_1 = K_2$), but it may be simplified further. Again, since $K_1 \subset K_2$ it suffices to show $K_2 \subset K_1$. To this end, take an arbitrary $k_2\in K_1$. Then $(e_H, k_2) \in H \times K_2 = H \times K_1$, which implies $k_2 \in K_1$. Therefore $K_2 \subset K_1$.

 

[stragequark]

Hi stragequark,

Your attempted proof is actually correct (you can just say at the end that since $K_2 \subset K_1$ as well, $K_1 = K_2$), but it may be simplified further. Again, since $K_1 \subset K_2$ it suffices to show $K_2 \subset K_1$. To this end, take an arbitrary $k_2\in K_1$. Then $(e_H, k_2) \in H \times K_2 = H \times K_1$, which implies $k_2 \in K_1$. Therefore $K_2 \subset K_1$.

Are you sure? This proof doesn't use any properties of subgroups or G being Abelian, which is why I'm worried.
I'm not too sure that the implication in
$(e_H, k_2) \in H \times K_2 = H \times K_1$, which implies $k_2 \in K_1$
is true.

 

[Euge]

Are you sure? This proof doesn't use any properties of subgroups or G being Abelian, which is why I'm worried.
I'm not too sure that the implication in
$(e_H, k_2) \in H \times K_2 = H \times K_1$, which implies $k_2 \in K_1$
is true.

Most of the assumptions you're worried about concern the existence, not uniqueness, of an internal direct product representation for $G$. The uniqueness property follows from equality of cartesian products (of nonempty sets): $A \times B = C \times D$ if and only if $A = C$ and $B = D$. So in fact, the hypothesis $K_1 \subset K_2$ is redundant. For given $G = H \times K_1$ and $G = H\times K_2$, we have $H \times K_1 = H \times K_2$. Since $H$, $K_1$ and $K_2$ are nonempty (being subgroups of $G$), $K_1 = K_2$ by equality of cartesian products.

 

[blue_raver22]

Your proof is correct and does not require the assumption that K1 and K2 are finite. The fact that G is Abelian and has two representations as the internal direct product of subgroups already implies that K1 and K2 must have the same order. This is because the order of G is equal to the product of the orders of its subgroups, and since G=HxK1 and G=HxK2, the orders of H and K1 must be equal, and the orders of H and K2 must be equal. Therefore, the orders of K1 and K2 must also be equal.

Your proof is also more general and does not rely on the assumption that K1 is a subset of K2. It works even if K1 and K2 are not subsets of each other, as long as they have the same order. This is because the product (h, k_1*k_2) is still an element of G=HxK1, and k_1*k_2 is still an element of K1, even if k_1 and k_2 are not in the same subgroup.

Overall, your proof is a good demonstration of the concept of direct product representations and how they can be used to show that two subgroups are equal. Keep up the good work!