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.

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

Answered: "Given the groups R× and Z, let G=R×Z...."

ankarskogC

Answered question

2021-02-03

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

Answer & Explanation

likvau

Skilled2021-02-04Added 75 answers

Here $G = R \times Z$ and the binary operation defined as follows
$(a, m) \times (b, n) = (a b, m + n)$
where $(a, m), (b, n) \in R \times Z$ .. Since $(1, 0) \in R \times Z$ , so G is non empty.
Associative: Let $(a, m), (b, n), (c, p) \in R \times Z$ ,
$(a, m) \times ((b, n) \times (c, p)) = (a, m) \times (b c, n + p) = (a b c, m + n + p) = (a b, m + n) \times (c, p) = ((a, m) \times (b, n)) \times (c, p)$ .
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 $(a, m) \in$
$(a, m) \times (1. a, - m) = (a \cdot 1. a, m - m) = (1, 0) = (1. a, - m) \times (a, m)$ . Hence, G is a group

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