Question

1. (a) Let (G, ∗) be a group. Prove that the map π : G −→ G defined by π(g) = g ∗ g is a
homomoprhism if and only if G is abelian.
(b) Show that the set
P =

a2t
2 + a1t + a0 |a2 + a1 = a0 and a2, a1, a0 ∈ R

is a group under addition.
(c) Consider the set X = R\{−1} with the binary relation ∗ defined by x ∗ y = x + y + xy.
Find the solution for the equation 5 ∗ x ∗ 2 = −19.

Answer 1

1(a)) Given that the map π : G → G defined by π(g) = g ∗ g.

$\pi(g_1*g_2) = (g_1*g_2)* (g_1*g_2) = g_1*(g_2*g_1)*g_2$ (because G is a group).

$= g_1*(g_1*g_2)*g_2$ iff $(G,*)$ is an abelian group

$\implies \pi(g_1*g_2) = (g_1*g_1)*(g_2*g_2) = \pi(g_1)*\pi(g_2)$

Thus, Given mapping is homomorphism if and only if G is abelian group.

1(b)) Given $P = \{a_2 t² + a_1 t + a_0 | a_2+ a_1 = a_0 \ and \ a_2, a_1, a_0 \in \R\}$ .

(i) Closure: Let A∈P,B∈P , so $A = a_2 t² + a_1 t + a_0 , B = a_2' t² + a_1' t + a_0'$ ​

and $a_2+a_1=a_0, a_2'+a_1'=a_0'$.

So, $(a_2+a_2')+(a_1+a_1') = a_0 +a_0'$ and A+B∈P . Hence, Closure property holds.

(ii) Associative: Also, (A+B)+C=A+(B+C) ∀ A,B,C∈P .

Hence, Associativity property holds.

(iii) Identity: ∃ 0∈P:0+A=A for all A∈P . Hence, 0 is identity exists in P.

(iv) Inverse: $\forall \ A\in P$, ∃ B=−A∈P:A+B=0 .

Hence, the set P is group under addition.

1(c)) Given  the set X = R\{−1} with the binary relation ∗ defined by x ∗ y = x + y + xy. 

Now, $5 ∗ x ∗ 2 = −19 \implies (5+x+5x)*2 = -19$

$\implies 5+x+5x+2+10+2x+10x = -19 \\ \implies 17+18x=-19 \\ \implies 18x = -36 \\ \implies x=-2$