Question

Let α : Z₉* Z₂₇→ Z₂₇ be given by α ((a, b)) = 3b for a ϵ Z₉, b ϵ Z₂₇ and
(a, b) + (c, d) = (a + c, b + d) is given by addition modulo n for each group Zn.
i. Show that α is a homomorphism.
ii. Find Ker(α).

Answer 1

Given $\alpha : Z₉ \times Z₂₇→ Z₂₇$ be given by $\alpha (a, b) = 3b$ for $a ϵ Z_9, b ϵ Z_{27}$ and

$(a, b) + (c, d) = (a + c, b + d)$ .

(i) Homomorphism: $\alpha((a, b) + (c, d)) = \alpha(a + c, b + d) = 2(b+d) = 2b + 2d$

Also, $\alpha(a,b)+\alpha(c,d) = 2b+2d$ .

Hence, $\alpha((a, b) + (c, d)) = \alpha(a, b) + \alpha(c, d)$ .

Hence, $\alpha$ is a Homomorphism map.

(ii) $ker(\alpha) = \{(a,b): \alpha(a,b)=0 \}$

$\alpha(a,b)=0 \implies 2b = 0 \implies b = 27 n$ where $n\in \Z$ .

So, $ker(\alpha) = \{(a,27n): n \in \Z\}$