MHB 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.