0
$\begingroup$

Prove $(H \times 1)(1\times K)= H\times K$ where $H,K$ are groups.

Suppose $x=ab,a\in H\times 1,b\in 1\times K$
Then $x=(h,1)(1,k)$ where $h\in H,k\in K$
Hence $x=(h,k)\in H\times K$

Let $(h,k)\in H\times K$
Then $(h,k)=(h,1)(1,k)\in (H\times 1)(1\times K)$

What I want to ask is $(h,1)(1,k)=(h,k)$ always true and is there any difference between $(H\times 1)(1\times K)$ and $(H\times 1)\times(1\times K)$?

$\endgroup$
0
$\begingroup$

Multiplication in $H\times K$ is given by $$(h_1,k_1)(h_2,k_2)=(h_1h_2,k_1k_2)$$ so, yes, $(h,1)(1,k)=(h1,1k)=(h,k)$.

Now, elements of $H\times K=(H\times 1)(1\times K)$ are pairs $(h,k)$ with $h\in H$ and $k\in K$. Elements of $(H\times 1)\times(1\times K)$ are pairs of pairs $((h,1),(1,k))$ with $h\in H$ and $k\in K$. Of course, there is an obvious isomorphism $$H\times K\to (H\times 1)\times (1\times K)$$ so this difference is superficial.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.