User:Michellepoliseno /Math735 MIDTERM

(1) Let G be a group. Consider the subset $$ H = {(x,x)| x \in G} \subset GxG. \ $$

Which of the following claims is true:

(1) H is a subgroup of G,

(TRUE)

Since G is a group, $$ e \in G \ $$ and thus by definition of H, $$ (e,e) \in H \ $$, and therefore H is not empty. Let $$ (x,x) \ $$ and $$ (y,y)^{-1} \in H \ $$. Then $$ (x,x)*(y,y)^{-1} = (x,x)*(y^{-1},y^{-1}) = (xy^{-1},xy^{-1}) \in H \ $$. Therefore H is a subgroup of G.

(2) H is a cyclic subgroup of G,

(3) H is always a normal subgroup of GxG,

(4) H is in general not a normal subgroup of GxG, but there are examples of noncommutative groups G such that H is a normal subgroup of GxG,

(5) H is a normal subgroup of GxG if and only if G is commutative?

(2) True of False:

(1) Every nontrivial group G contains a nontrivial proper subgroup H<G,

(FALSE)

Suppose that G is finite and of prime order $$ P \ $$. Then Lagrange's Theorem implies that any subgroup must divide $$ P \ $$. By definition of prime, any subgroup of $$ P \ $$ has an order of $$ 1 \ $$ or $$ P \ $$. Hence G can only have itself and the Trivial Group as subgroups.

(2) Every nontrivial group contains a nontrivial normal proper subgroup,

(3) Every group H can be embedded (i.e. mapped by an injective homomorphism) into some group G so that G is larger than the image of H and the image of H is normal in G.

(3) Let $$ f:R \implies S \ $$ be a homomorphism of unitary commutativite rings and let $$ I \subset R \ $$ be an ideal contained in the ker(f). Show $$ \exists! \ $$ ring homomorphism $$ g:R/I \implies S \ $$ for which $$ g \cir \pi = f \ $$ where $$ \pi : R \implies R/I \ $$ is the 'canonical' homomorphism, i.e., $$ \pi(a) = (\overline{a}) \ $$ (Notice: the problem has the existence and the uniqueness parts.)