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,

(FALSE) H is not a subgroup of G, since in order to be a subgroup H must be a subset of G, but H is not a subset of G.

H is, however, a subgroup of GxG.

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 $$ (y,y) \in H \ $$ for some $$ x, y \in G \ $$. Since we found that $$ (e,e) \in H \ $$ we can say $$ (y,y)^{1} \in H \ $$. Then $$ (x,x) \ $$ x $$ (y,y)^{-1} = (x,x) \ $$ x $$ (y^{-1},y^{-1}) = (xy^{-1},xy^{-1}) \in H \ $$. Therefore H is a subgroup of GxG.

(2) H is a cyclic subgroup of G,

(FALSE) Since H is not a subgroup of G, it cannot be a cyclic subgroup of G.

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

(FALSE)

(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,

(FALSE)

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

(FALSE)

(2) True of False:

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

(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.

(FALSE)

Let $$ f: H \to G \ $$ be an injective homomorphism.

(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 $$ g \circ \mathcal \pi = f \ $$ 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.)

Existence: Let $$ a,b \in R \ $$. Note since $$ \pi : R \implies R/I \ $$ is a homomorphism, then $$ \pi (ab) = \pi (a) \pi (b) = (\overline{a})(\overline{b}) \ $$ for some $$(\overline{a}) \ $$ and $$ (\overline{b}) \in R/I \ $$. And $$ f(ab) = f(a)f(b) \ $$ for some $$ f(a), f(b) \in S \ $$.

Then, $$ g(\overline{a} \overline{b}) = f(ab) \ $$, by definition of f. And $$ f(ab) = f(a)f(b) \ $$, since f is a homomorphism. Then, by definition of f, $$ f(a)f(b) = g(\overline{a})g(\overline{b}) \ $$, where $$ g(\overline{a}), g(\overline{b}) \in S \ $$. Thus g is a homomorphism.

Uniqueness:

(4) In the list below, identify all mutually isomorphic pairs of rings:

A = $$ \C [X]/(X^{2}) \ $$

B = $$ \C [X]/((X-1)^{2}) \ $$

C = $$ \C [X]/(X^{3}) \ $$

D = $$ \C [X]/(X^{2}+1) \ $$

E = $$ \C X \C \ $$

(5) In the list of rings in Problem (4), identify all mutually isomorphic pairs of F-vector spaces. (Notice: here we mean the naturally existing F-Vector space structures on any ring containing an isomorphic copy of F as a subring.)

(6) Let F be a field of positive characteristic $$ p>0 \ $$. In one of the homework assignments it was shown that the map $$ F \to F \ $$, $$ a \to a^{p} \ $$, is a ring homomorphism. Show that: (1) If F is finite, then the mentioned map is an automorphism of F. (2) Does part (1) extend to the general case when F is an arbitrary (not necessarily finite) field of characteristic $$ p>0 \ $$?