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.