User:MCPOliseno /Math850 HW1
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) \ $}. Thus every nonidentity element of $ G - t(G) \ $ has order 2.
1.6.8) Prove that is $ n \ne m \ $ then $ S_n \ $ and $ S_m \ $ are not isomorphic.
Finite groups of different order cannot be isomorphic. Thus $ S_n \ $ and $ S_m \ $ are not isomorphic.
1.6.9) Prove that $ D_{24} \ $ and $ S_4 \ $ are not isomorphic.
(In $ D_{24} \exists \ $ order 12 that is not in $ S_4 \ $)
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 \ $.
Since, $ H \ $ is a subset of $ \Q \ $ $ H \ $ is closed under addition, then $ \frac{a}{b} +\frac{a}{b} \dots + \frac{a}{b} \ $ b times is still in $ H \ $. Thus $ a \in H \implies \frac{1}{a} \in H \ $. Similarly, $ \frac{1}{a} + \frac{1}{a} + \dots + \frac{1}{a} \ $ can be written a times, such that $ 1 \in H \ $ and therefore $ \Z \subseteq H \ $. Then $ \forall x \in \Z, \frac{1}{a} \in H \implies H = \ $ {$ \frac{a}{b} : a, b \in \Z \ $} which is $ \Q \ $ and thus $ H = \Q \ $.
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 \ $.
$ \sigma_a \ $ is an automorphism $ \iff \ $ it is surjective
$ \iff \ $ $ f^{-1} (1) = x \ $ $ \iff \ $ $ ax = 1 \ $ $ \iff \ $ $ ax - 1 = bn \ $ $ \iff \ $ $ ax - bn = 1 \ $ $ \iff \ $ $ gcd (a,n) = 1 \ $
(b) Prove that $ \sigma_a = \sigma_b \iff a \equiv b(mod n) \ $.
$ \sigma_a = \sigma_b \iff ax = bx \ $ $ \iff \ $ $ ax-bx = 0 \ $ $ \iff \ $ $ x(a-b) = 0 \ $ $ \iff \ $ $ a-b = nz \ $ for some $ z \in \Z \ $ $ \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 \ $ is an isomorphism of $(\Z/n\Z)^x \ $ onto the automorphism group of $\Z_n \ $.
2.4.14 c) A group $ H \ $ is called finitely generated if there is a finite set $ A \ $ such that $ H = <A> \ $. Prove that every finitely generated subgroup of the additive group $ \Q \ $ is cyclic.
Let $ H \ $ be finitely generated subgroup of $ \Q \ $. Then $ H = <a_1, \dots, a_n> \ $ 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.