Let $H$ and $ K$ be groups and let $H \times 1 = \{ (h, 1_K) | h \in H\}$ and $1 \times K = \{(1_H, k) | k \in K\}$, where $1_H$ and $1_K$ denote the identity elements of $H$ and $K$ respectively.
Prove that,
(i) Every element of $H \times1$ commutes with every element of $1\times K$ and $H \times 1 \cap1 \times K = (1_H, 1_K)$.
Attempt,
take any $(h,1_K) \in H \times 1$ and $(1_K,k) \in 1 \times K$, then $$(h,1_K).(1_H,k)=(h1_H,1_Kk)=(h,k)$$
$$(1_H,k).(h,1_K)=(1_Hh,k1_K)=(h,k)=(h,1_K).(1_H,k)$$ $ \therefore$ Every element of $H \times1$ commutes with every element of $1\times K$
For the second part,
It clear that $(1_H,1_K) \in H \times 1$ and $(1_H,1_K) \in 1 \times K$ this follows $(1_H,1_K) \in H \times 1 \cap 1 \times K$
Suppose $ a \in H \times 1 \cap 1 \times K$ and $a \not= (1_H,1_K)$,
This follows $a \in H \times 1$ and $ a \in 1 \times K$ hence by definition of $H \times 1$ and $ 1 \times K$ ,
$a=(h_1,1_K)$ and $b=(1_H,k_1)$ where $h_1 \in H$ and $k_1 \in K$.
this follows $(h_1,1_K)=(1_H,k_1) $ so $h_1=1_H$ and $k_1=1_K$
which contradicts $a \not= (1_H,1_K)$,That is $H \times 1 \cap1 \times K=(1_H, 1_K)$
Can anyone verify my answer? Second part is not sure.