Question

# Given the groups R∗ and Z, let G = R∗×Z. Define a binary operation ◦ on G by (a,m)◦(b,n) = (ab,m + n). Show that G is a group under this operation.

Analyzing categorical data

Given the groups $$R\times$$ and Z, let $$G = R×Z$$. Define a binary operation $$\circ$$ on G by $$(a,m)\times(b,n) = (ab,m + n)$$. Show that G is a group under this operation.

2021-02-04

Here $$\displaystyle{G}={R}×{Z}$$ and the binary operation defined as follows
$$(a,m)\times(b,n)=(ab,m+n)$$
where $$(a,m),(b,n)\in R\times Z$$.. Since $$(1,0)\in R\times Z$$, so G is non empty.
Associative: Let $$(a,m),(b,n),(c,p)\in R\times Z$$,
$$(a,m)\times((b,n)\times(c,p))=(a,m)\times(bc,n+p)=(abc,m+n+p)=(ab,m+n)\times(c,p)=((a,m)\times(b,n))\times(c,p)$$.
$$(a,m)\times((b,n)\times(c,p)) =(a,m)\times(bc,n+p) =(abc,m+n+p) =(ab,m+n)\times(c,p) =((a,m)\times(b,n))\times(c,p)$$.
Identity: Claim (1,0) is the identity of G, since for any $$(a,m)\in G(a,m)\in G$$ we have
$$(a,m)\times(1,0)=(a,m)=(1,0)\times(a,m)$$.
Inverse:
For any $$(a,m)\in$$
$$(a,m)\times(1.a,−m)=(a⋅1.a,m−m)=(1,0)=(1.a,−m)\times(a,m)$$. Hence, G is a group​