Let $(\Bbb{Z}_6,+)$ and $(\Bbb{Z}_5,\cdot)$ be two groups.
Define an operation for $(\Bbb{Z}_6,+)\times(\Bbb{Z}_5,\cdot)$ such that $(\Bbb{Z}_6,+)\times(\Bbb{Z}_5,\cdot)$ is a group. Find the identity and inverses of the product.
I know the following:
Property If $(G_1,*_1)$ and $(G_2,*_2)$ are two GROUPS, then the set $G_1\times G_2$ with the operation $(a,b)*(c,d)=(a*_1c,b*_2d)$ is also a GROUP.
But here, $(\Bbb{Z}_5,\cdot)$ is NOT a group since the inverse element of $0$ does not exist. So what should we do here?
**Should we consider another operation like $$(a,b)*(c,d)=(a+_6c,b+_5d)?$$
But the hypothesis is not true because $(\Bbb{Z}_5,\cdot)$ is not a group, so I do not know if I can define such operation of the product of "two" groups.
Any help?
Assuming $(\Bbb Z_5,\cdot)=U(5)$ is the group of units modulo five, the identity of $G:=(\Bbb Z_6, +)\times U(5)$ is $$([0]_6, [1]_5)$$ and the inverse of $([a]_6, [b]_5)$ in $G$ is $([-a]_6, [c]_5)$, where $c$ is such that $5x+bc=1$ for some $x\in \Bbb Z$ and $$[d]_n:=\{k\in \Bbb Z\mid n \text{ divides } (k-d)\}.$$
Assuming $(\Bbb Z_6,+)=(\Bbb Z_6, +_6)$ and $(\Bbb Z_5,\cdot)=(\Bbb Z_5, +_5)$, then the identity of $H:=(\Bbb Z_6, +_6)\times(\Bbb Z_5, +_5)$ is $$([0]_6, [0]_5)$$ and the inverse of $([a]_6, [b]_5)$ in $H$ is $([-a]_6, [-b]_5)$.
NB: Since five is prime, there is only one group of order five up to isomorphism. The question specifies that $(\Bbb Z_5,\cdot)$ is a group.