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.

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.

Question
Analyzing categorical data
asked 2021-02-03

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.

Answers (1)

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​

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