User:MCPOliseno /Math850 HW1

1.1.25) Prove that if $$ x^2 \ $$ = 1 $$ \forall x \in G \ $$ then $$ G \ $$ is abelian.

Let $$ a, b \in G \ $$. Then $$ ab = a * 1 * b \ $$

= $$ a (ab)^2 b \ $$, since $$ x^2 \ $$ = 1

= $$ a(ab)(ab)b \ $$

= $$ (aa)(ba)(bb) \ $$

= $$ a^2(ba)b^2 \ $$

= $$ 1(ba)1 \ $$, since $$ x^2 \ $$ = 1

= $$ ba \ $$

Thus $$ ab = ba \ $$ and therefore $$ G \ $$ is abelian.

1.1.31) Prove that any finite group of $$ G \ $$ of even order contains an element of order 2.

Let $$ t(G) \ $$ = {$$ g\in G| g \ne g^{-1} \ $$}. We want to show that $$ t(G) \ $$ has an even number of elements and that every nonidentity element of $$ G - t(G) \ $$ has order 2.

Note that {$$ t(G) \ $$} = {$$ g_1, g_1^{-1} \ $$} $$ \cup \ $$ {$$ g_2, g_2^{-1} \ $$} $$ \cup \dots \ $$ $$ \cup \ $$ {$$ g_k, g_k^{-1} \ $$}. Then clearly $$ |t(G)| \ $$ is even.

Then $$ |G - t(G)| \ $$ = $$ |G| - |t(G)| \ $$, which is evidently even. Thus 2 divides $$ |G| \ $$ and 2 divides $$ |t(G)| \ $$ and therefore 2 divides $$ |G - t(G)| \ $$. Note that $$ 1 \in G and 1 \notin t(G) \ $$ which implies that $$ 1 \in \ $$ {$$ G - t(G) \ $$}.

1.6.8) Prove that is $$ n \ne M \ $$ then $$ S_n \ $$ and $$ S_m \ $$ are not isomorphic.

(the order is not the same)

1.6.9) Prove that $$ D_{24} \ $$ and $$ S_4 \ $$ are not isomorphic.

(the order is not the same)

2.1.13) Let $$ H \ $$ be a subgroup of the additive group of rational numbers with the property that $$ 1/x \in H \ $$ for every nonzero element $$ x \in H \ $$. Prove that $$ H = 0 \ $$ or $$ H = \Q \ $$.

$$ \to \ $$ Suppose that $$ H \ne 0 \ $$. We want to show that $$ H = \Q \ $$.

$$ \gets \ $$ Suppose that $$ H\ne \Q \ $$. We want to show that $$ H = 0 \ $$.

2.1.17) Let $$ n \in \Z^+ \ $$ and let $$ F \ $$ be a field. Prove that the set $$ G \ $$ = {$$(a_{ij}) \in GL_n(F) | a_{ij} = 0 \forall i > j, a_{ii}=1 \forall i \ $$} is a subgroup of $$ GL_n(F) \ $$.

Let $$ A, B \in G \ $$. Then $$ A = \begin{bmatrix} 1 & a_{12} & \cdots & a_{1j} \\ 0 & 1 & \cdots & a_{2j} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{bmatrix} \ $$ And $$ B = \begin{bmatrix} 1 & b_{12} & \cdots & b_{1j} \\ 0 & 1 & \cdots & b_{2j} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{bmatrix} \ $$. Then $$ AB = \begin{bmatrix} 1 & c_{12} & \cdots & c_{1j} \\ 0 & 1 & \cdots & c_{2j} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{bmatrix}\ $$ where $$ c_{ij} = a_{ij}b_{ij} \ $$. Thus $$ AB \in G \ $$. Now $$ A^{-1} \ $$ = $$ \begin{bmatrix} 1 & 1/a_{12} & \cdots & 1/a_{1j} \\ 0 & 1 & \cdots & 1/a_{2j} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{bmatrix} \ $$. And thus $$ A^{-1} \in G \ $$ and therefore $$ G \ $$ is a subgroup of $$ GL_n(F) \ $$.

2.3.26) Let $$ \Z_n \ $$ be a cyclic group of order $$ n \ $$ and for each integer $$ a \ $$ $$ \sigma_a : \Z_n \to \Z_n \ $$ by $$ \sigma_a (x) = x^a \ $$ for all $$ x \in \Z_n \ $$.

(a) Prove that $$ \sigma_a \ $$ is an automorphism of $$ \Z_n \iff gcd(a,n) = 1 \ $$.

$$ \to \ $$ Assume that $$ \sigma_a \ $$ is an automorphism and that $$ gcd(a,n) = k \ $$, where $$ k \ne 1 \ $$. Note $$ gcd(a,n)= k \implies k|a \ $$ and $$ k|n \implies a = kl, n = km \ $$ for some $$ l,m \in \Z \ $$. Then $$ k = n/m \implies a = nl/m \ $$. Then $$ \sigma_a (x) \ $$ = $$ x^a \ $$ = $$ x^{nl/ m} \ $$ = $$ (x^n)^{l/ m} = (0)^{l/m} = 0 \ $$. Since $$ \Z_n \ $$ is cyclic order $$ n \ $$. This is a contradiction of our supposition since this implies that $$ \sigma_a \ $$ is not bijective and therefore could not be an automorphism. It follows then, that $$ gcd(a,n) = 1 \ $$.

(b) Prove that $$ \sigma_b = \sigma_b \iff a \equiv b(mod n) \ $$.

(c) Prove that every automorphism of $$ \Z_n \ $$is equal to $$ \sigma_a \ $$ for some integer $$ a\ $$.

(d) Prove that $$ \sigma_a \circ \sigma_b = \sigma_{ab} \ $$. Deduce that the map $$ \overline{a} \to \sigma_b \  \. Prove that every finitely generated subgroup of the additive group $$ \Q \ is cyclic.

Let $$ H \ $$ be finitely generated subgroup of $$ \Q \ $$. Then $$ H =  \ $$ where $$ a_i \in \Q, 1 \le i \le n \ $$. So $$ a_i = \frac{p_i}{q_i} \ $$ for $$ p_i, q_i \in \Z \ $$ and $$ 1 \le i \le n \ $$. We want to show that $$ H \subset <\frac{1}{k}> \ $$ where $$ k \ $$ is the product of the denominators of the $$ a_is \ $$/. So let $$ g \ in H \ $$. Then $$ g = \alpha_1 a_1 + \dots + \alpha_n a_n \ $$ where $$ \alpha_i \in \Z, 1\le i \le n \ $$. Then $$ g = \alpha_1 \frac{p_1}{q_1} + \dots + \alpha_i \frac{p_i}{q_i} = \sum_{i=1}^{n} \alpha_i p_n (q_1 \dots q_n) / (q_1 \dots q_n) \ $$. Then $$ g \in <\frac{1}{k}> \ $$ where $$ k = q_1 \dots q_n \ $$. We know that any subgroup of a cyclic subgroup is cyclic. Therefore $$ H \ $$ is cyclic.