User:J D Bowen/Math735 HW3

1) Let $$F \ $$ be a field of finite order $$q \ $$, and $$n\in\mathbb{Z}_+ \ $$.

Observe that the determinant is a homomorphism $$\Delta:GL_n(F)\to F \backslash \left\{{0}\right\} \ $$. Then by the corollary to the first isomorphism theorem,

$$[GL_n(F):\text{ker}(\Delta)] = |\Delta(GL_n(F))| \ $$.

But $$\text{ker}(\Delta) = SL_n(F) \ $$ and $$\Delta(GL_n(F))=F\backslash\left\{{0}\right\} \ $$.

So $$[GL_n(F):SL_n(F)]=q-1 \ $$.

3.3.4) Clearly $$C\times D < A \times B \ $$. We have $$(a,b)(C,D)=(aC,bD)=(Ca,Db)=(C,D)(a,b) \ $$.

Define $$\phi:(A\times B)/(C\times D) \to A/C \times B/D \ $$ as $$\overline{(a,b)}\mapsto (\overline{a},\overline{b}) \ $$.

Then $$\phi(\overline{(a,b)}\cdot\overline{(x,y)} = \phi(\overline{(ax,by)}=(\overline{ax},\overline{by})=(\overline{a},\overline{b})\cdot(\overline{x},\overline{y}) = \phi(\overline{(a,b)})\cdot\phi(\overline{(x,y)}) \ $$, so $$\phi \ $$ is a homomorphism.

We also have $$\text{ker}(\phi) = \left\{{ (a,b)(C\times D) : \phi((a,b)(C\times D)) = 1 \cdot C\times D }\right\} \ $$

$$=\left\{{ (a,b)(C\times D) : aC\times bD = 1 \cdot C\times D }\right\} = \left\{{(A,b)(C\times D): (a,b)\in C\times D }\right\} = C\times D \ $$.

7.1.29) Let $$R=\left\{{\phi:A_+\to A_+: \phi \ \text{is hom.} }\right\} \ $$.

We easily see for $$\alpha,\beta\in R, \alpha+\beta=\beta+\alpha\in R \ $$. We also have the map $$0_A \ $$ defined $$x\mapsto 0 \ $$, for an identity element. Since composition is associative, and compositions of homomorphisms is another homomorphism, composition and pointwise addition give $$R \ $$ a ring structure.

Suppose $$\phi\in U(R) \ $$. Then $$\phi:A_+\to A_+ \ $$ is a homomorphism and so is $$\phi^{-1}:A_+\to A_+ \ $$ is also homomorphism. Therefore, $$\phi \ $$ is one-to-one, which implies $$\phi \ $$ is an automorphism.

Now suppose $$\phi\in\text{Aut}(A) \ $$. Then $$\phi \ $$ is 1-1 and a homomorphism. So there is a homomorphism $$\phi^{-1} \ $$ such that the composition is the identity map. Then $$\phi\in U(R) \ $$.

So $$U(R)=\text{Aut}(A) \ $$.

PROOF PHI INVERSE IS HOM:

Since $$\phi \ $$ is 1-1, exists unique inverse $$\phi^{-1} \ $$.

$$a+b= \ $$

$$\phi(\phi^{-1}(a)) + \phi(\phi^{-1}(b)) = \ $$

$$\phi(\phi^{-1}(a)+\phi^{-1}(b)) \ $$, since $$\phi \ $$ is a homomorphism.

Now we can $$\phi^{-1} \ $$ both sides of this, to get

$$\phi^{-1}(a+b) = \phi^{-1}(a)+\phi^{-1}(b) \ $$

so the inverse is a homomorphism, too.