Answer to Question #123756 in Abstract Algebra for Felix Coleman

Question #123756
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.
1
Expert's answer
2020-06-28T16:43:31-0400

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 t\u00b2 + a_1 t + a_0 | a_2+ a_1 = a_0 \\ and \\ a_2, a_1, a_0 \\in \\R\\}" .

(i) Closure: Let AP,BP , so "A = a_2 t\u00b2 + a_1 t + a_0 , B = a_2' t\u00b2 + 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+BP . Hence, Closure property holds.

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

Hence, Associativity property holds.

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

(iv) Inverse: "\\forall \\ A\\in P", ∃ B=−AP: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 \u2217 x \u2217 2 = \u221219 \\implies (5+x+5x)*2 = -19"

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


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