User:J D Bowen/Math735 HW2

1.6


 * 17.

Suppose the map from G to itself defined by $$\phi(g)=g^{-1} \ $$ is a homomorphism. Then by the definition of homomorphism we know that $$\forall a,b\in G, \phi(ab)=\phi(a)\phi(b) \ $$. However, according to the map, $$\phi(ab)=(ab)^{-1}=b^{-1}a^{-1} \ $$ and $$\phi(ab)=\phi(a)\phi(b)=a^{-1}b^{-1} \ $$, so $$\phi(a)\phi(b)=\phi(b)\phi(a) \ $$. Hence, G is abelian.

Now suppose G is abelian. Then $$\phi(a)\phi(b)=\phi(b)\phi(a) \ $$. We know that our map is defined as $$\phi(ab) = (ab)^{-1} \ $$. Since G is abelian, $$(ab)^{-1}= b^{-1}a^{-1} = a^{-1}b^{-1} = \phi(a)\phi(b) \ $$. Therefore, $$\phi(ab)=\phi(a)\phi(b) \ $$, which means the map from G to itself is a homomorphism.

. Prove that for any fixed integer k < 1, the map from G to itself defined by
 * 1) 19. Let G =

is a surjective homomorphism but is not an isomorphism. Let z1 and z2 be elements of G. Then ϕ( z1 z2) = (z1 z2)k = z1k z2k = ϕ(z1)ϕ(z2). Therefore, the map is a homomorphism. Now consider the element zk in the image of G. Since G is being

mapped to itself, there exists some

that for some . Therefore, , so z is an element of G. Thus, the map is a surjective homomorphism. Next, to prove that the map is not an isomorphism, we need to show that it is not injective. Consider the elements . Clearly, . Now set , and observe that ϕ(1) = 14 = 1 and ϕ(i) = i4 = 1. Therefore, ϕ(1) = ϕ(i). Thus the map is not injective, and not an isomorphism.

Let $$p \ $$ be prime, $$n\in\mathbb{N} \ $$, and suppose $$\exists x \in G \ $$ such that $$x^{(p^n)} =1 \ $$. We aim to show that $$|x|=p^m \ $$ for some $$m\leq n \ $$.

We have $$x^{(p^n)} =1\implies |x|<\infty \implies H:=\langle x \rangle \ $$ is a finite subgroup. This, combined with $$x^{(p^n)}=1 \ $$ implies $$|H| \mid p^n \implies \exists q\leq n \ $$ such that $$|H|=p^q \ $$, since $$p \ $$ is prime. Since $$x\in H \ $$ and $$|H|<\infty, |x| \mid |H| \implies |x| \mid p^q \implies \exists m\leq q \leq n \ $$ such that $$|x|=p^m \ $$.

Let $$h\in H \ $$. We aim to show there is a unique homomorphism $$\phi:\mathbb{Z}\to H \ $$ such that $$1\mapsto h \ $$. Define $$\phi(n)=h^n \ $$. Then $$\phi(a+b)=h^{a+b}=h^a h^b = \phi(a)\phi(b) \ $$, so $$\phi \ $$ is a homomorphism. Now let $$\phi \ $$ be any homomorphism fitting the requirements. Then for any element $$n\in\mathbb{Z} \ $$, we have

$$\phi(n)=\phi(\underbrace{1+1+\dots+1}_{n \ \text{times}}) = \underbrace{\phi(1)\cdots \dots\cdot\phi(1)}_{n \ \text{times}} = \underbrace{h\cdot h \cdot \dots h}_{n \ \text{times}} = h^n \ $$

so the homomorphism is unique.

Let $$A=(a_{ij}) \in GL_n(F) \ $$ such that $$i<j\implies a_{ij}=0 \ $$. We wish to demonstrate all such matrices form a group. So let $$B \ $$ be a similarly defined matrix, and observe that

$$(AB)_{ij} = \sum_{k=1}^n a_{ik}b_{kj} \ $$.

If this entry is not zero, it must be that both $$i\geq k, k\geq j \ $$. Taking the contrapositive, if $$i<j, (AB)_{ij}=0 \ $$. So multiplication in this set is closed.

Now let $$B \ $$ be any matrix, $$A \ $$ defined as before, and suppose $$AB=I \ $$. Then

$$\sum_{k=1}^n a_{ik}b_{kj} = \delta_{ij} \ $$,

but since $$a_{ij}=0 \ $$ for $$i<j \ $$, this is just

$$\delta_{ij}=\sum_{k=1}^n a_{ik}b_{kj} = \sum_{k= i}^n a_{ik}b_{kj} \ $$.

This means we can allow $$b_{ij}=0 \ $$ for $$i<j \ $$ and get the same result. Hence, $$A^{-1} \ $$ is in the set as well.

We aim to show that $$\mathbb{Q}^+ = \langle \left\{{1/p : p \ \text{prime} }\right\} \rangle \ $$.

Let $$x=a/b \in \mathbb{Q}^+ \ $$. By the fundamental theorem of arithmetic, there exists a unique prime decomposition of $$a,b \ $$ such that

$$x=\frac{a}{b} = \frac{p_{x,1}^{\alpha_{x,1}} p_{x,2}^{\alpha_{x,2}} \dots p_{x,m}^{\alpha_{x,m}}} { p_{y,1}^{\alpha_{y,1}} p_{y,2}^{\alpha_{y,2}} \dots p_{y,n}^{\alpha_{y,n}}} \ $$

But of course the denominator here is certainly a product of powers of inverse primes, and so is in the group generated by the inverse primes, and since the inverses of the inverse primes are just primes, so is the numerator. Hence, we can construct any rational from elements of the group generated by the inverse primes.

Now consider any element $$z\in\langle \left\{{1/p : p \ \text{prime} }\right\} \rangle \ $$. Then $$z=p_1^{\alpha_1} \dots p_1^{\alpha_n} \ $$, where the alphas may be positive or negative. The primes with positive powers will uniquely determine some integer, and the primes with negative powers will uniquely determine the inverse of some integer, and so z must be in $$\mathbb{Q}_+ \ $$.

2.4 19. a) Prove that the additive group of rationals is divisible.

Let

. Since k is nonzero, we can divide it out to get . We have defined b and k to be nonzero, therefore x is a rational number and thus, the additive group of rationals is divisible.

b) Assume G is a divisible finite abelian group with |G| = k, for some non-zero integer k.

Let a ∈ G, where a does not equal the identity element e. Then, since G is divisible,

there exists an element x ∈ G, such that

xk = a. Then let |x|=m for some m integer. By Lagrange Theory, m divides k, so k=md, for some d integer. Then

xk = xmd = (xm)d = e ≠a. This contradicts our assumption, thus we can conclude that there is no finite group G that is divisible.