User:MCPOliseno /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)

Let $$ (a,b) \in \ $$ GxG, for some $$ a \ne b \in G \ $$ and $$ (y,y) \in H \ $$ for some $$ y \in G \ $$. Since GxG is a group, $$ (a,b)^{-1} \in \ $$ GxG, for some $$ a^{-1}, b^{-1} \in G \ $$. So $$ (a,b)(y,y)(a,b)^{-1} = (ay,ay)(a^{-1},b^{-1}) = (aya^{-1},byb^{-1}) \ $$. Since $$ a \ne b \ $$, then $$ (aya^{-1},byb^{-1}) \ $$ not in H. Thus H is not 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,

(FALSE)

Since (5) proves that H is only a normal subgroup of GxG if and only if G is commutative, then there cannot exist examples of noncummutative 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?

(TRUE)

$$ \rightarrow \ $$ Suppose H is a normal subgroup of GxG. We want to show that G is commutative. Since H is a normal subgroup of GxG, then $$ (axa^{-1}, bxb^{-1}) = (x, x) \in H \ $$, for some $$ (a, b) and (a, b)^{-1} \in \ $$ GxG, where $$ a, b, a^{-1}, b^{-1}, and x \in G \ $$. Now multiply (a, b) on the right of both sides of the equation:

$$ (axa^{-1}, bxb^{-1})(a, b) = (x, x)(a, b) \ $$

Then $$ (axa^{-1}a, bxb^{-1}b) = (xa, xb) \ $$

$$ \implies (ax, bx) = (xa, xb) \ $$. Thus ax = xa and bx = xb, and therefore G is commutative.

$$ \leftarrow \ $$ Suppose that G is commutative. Then:

$$ (axa^{-1}, bxb^{-1}) = (aa^{-1}x, bb^{-1}x) = (x, x) \in H \ $$. Thus H is normal in GxG.

(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,

(FALSE)

As shown in (1), $$ \exists \ $$ a nontrivial group G that is finite and of order $$ p \ $$, which contains only the trivial group and itself as subgroups and therefore does $$ not \ $$ contain a nontrivial proper subgroup, and thus cannot contain 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.

(TRUE)

Let H and G be groups and $$ f: H \to G \ $$, where $$ \forall h \in H, f \ $$ maps h to $$ e \in G \ $$, where e is the identity element in G. Then G is larger than the image of H and the image of H is normal in G.

We want to show that $$ f \ $$ is a homomorphism. So, let a, b $$ \in H\ $$. Then f(a) = e and f(b) = e. Then f(ab) = e = e*e = f(a)f(b). Thus $$ f \ $$ is a homomorphism. The Kerf = {h $$ \in H \ $$ | f(h) = e}, thus the Kerf = e. Now, suppose f(x) = f(y) $$ \implies f(xy^{-1}) = f(x)f(y)^{-1} \ $$ = e', where e' is the identity element of H. This implies that $$ xy^{-1} \in \ $$ kerf. Thus $$ xy^{-1} = e \ $$, which implies that x = y. Therefore $$ f \ $$ is injective. Hence every group H can be embedded 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 $$ g \circ \mathcal \pi = f \ $$ 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:

Assume the contrary, that $$ \exists \phi R/I \to S \ $$ such that $$ \phi \ne \ $$ g, where $$ \phi \circ \mathcal \pi = f \ $$. Then $$ \exists \ $$ a coset $$ \overline{x} \in R/I $$ such that $$ \phi (\overline{x}) \ne g(\overline{x}) \ $$.

Then f(x) = $$ \phi \circ \mathcal \pi (x) = \phi (\pi (x)) = \phi (\overline{x}) \ne g(\overline{x}) = g (\pi (x)) = g \circ \mathcal \pi (x) \ $$ = f(x). Which is a contradiction. And thus g is unique.

(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 = $$ \C X \C \ $$

C = $$ \C [X]/(X^{3}) \ $$, is not isomorphic to A, B, D or E. $$ \C [X]/(X^{2}+1) \ $$ = $$ \C [X]/(x+i) \ $$x$$ \C [X]/(x-i) \ $$, by Theorem in 9.5, $$ \C [X]/(x+i) \cong \C \ $$ and $$ \C [X]/(x-i) \cong \C $$, thus $$ \C [X]/(x+i) \ $$x$$ \C [X]/(x-i) \cong \C $$x$$ \C \ $$. A = $$ \C [X]/(X^{2}) \cong \ $$ B = $$ \C [X]/((X-1)^{2}) \ $$, since are both of order 2.

(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.)

By the Euclidean Algorithm, every polynomial p(x) can be written in the form p(x) = a(x)q(x) + r(x), where r(x) $$ \in F[X] \ $$ and 0 $$ \le \ $$ deg r(x) $$ \le \ $$ n-1. Any two finite dimensional vector spaces over F of the same dimension are isomorphic. Looking at A, B, D and E, all three of these have the same dimension (dimension 2). And thus A $$ \cong \ $$ B $$ \cong \ $$ D $$ \cong \ $$ E. The dimension of C, however, is 3, which is not equal to any of the others on the list, and thus C is not isomorphic to A, B, D or E.

(6) Let F be a field of positive characteristic $$ p>0 \ $$. In one of the homework assignments it was shown that the map $$ \sigma: F \to F \ $$, $$ a \to a^{p} \ $$, is a ring homomorphism.

Show that: (1) If F is finite, then the mentioned map is an automorphism of F.

Let F be finite. Since $$ ker \sigma = \ $$ {x $$ \in \ $$ F | $$ \sigma \ $$(x) = e} and $$ \sigma \ $$ is a homomorphism, which implies that $$ \sigma (e) = e \ $$, then $$ ker \sigma \ $$ = {e}, and thus $$ \sigma \ $$ is injective. An injective map from a finite set to itself must be bijective, and thus $$ \sigma \ $$ is bijective and therefore an automorphism of F.

(2) Does part (1) extend to the general case when F is an arbitrary (not necessarily finite) field of characteristic $$ p>0 \ $$?

$$ \sigma: F \to F \ $$, $$ a \to a^{p} \ $$ is a ring homomorphism and is always injective, but is not always surjective. The finiteness of F makes it bijective and therefore a automorphism, without F being finite, that won't be the case.