Problems

Problem 1

Is the function f \colon \mathbb{N} \to \mathbb{N}, f(x) = x + 1 surjective?

Yes, let n = x + 1.

Problem 2

Let G be a finite group such that the order of G is equal to n > 2. Prove G cannot have a subgroup H of order n - 1.

This can proved using Lagranges theorem. Towards a contradiction, assume there exists a subgroup H or G with order n-1. By lagranges theorem, the order of H must divide the order of G. This means

\exists n \in \mathbb{N}, n > 2, \; n - 1 \mid n

Which is a contradiction.

Problem 3

Let n \geq 3. Prove D_{2n} has a normal subgroup \left<r\right>.

Problem 5

Let [42] \in \mathbb{Z}/180\mathbb{Z}. Find |[42]|.

Because \gcd(42,180) = 6, and |x| = \frac{n}{\gcd(x,n)}. Thus |[42]| = 30.

Problem 10

Let GL_n(R) be the group of (n\times n) matrices with non-zero determinates. Let SL_n(R) be the group of (n \times n) matrices with determinate 1. Prove that SL_N(R) \leq GL_n(R)

Note that SL_n(R) is a subset of GL_n(R), because for every A \in SL_n(R), \det(A) = 1 \not = 0, thus A \in GL_n(R). We know SL_n(R) is not empty as the (n \times n) identity matrix has determinate 1.

Let A,B \in SL_n(R). Note that AB^{-1} has determinate 1. through algebra

\det(AB^{-1}) = \det(A)\det(B^{-1}) = 1 \not = 0

Thus AB^{-1} \in SL_n(R). Thus by subgroup criterian, we have shown SL_n(R) \leq GL_n(R).

Note that SL_n \trianglelefteq GL_n(R). To see this, take L \in Ln(R) with \det(L) = x. Thus for any \mathcal{o} \in SL_n(R),

\det(LoL^{-1}) = \det(L)\det(o)\det(L^{-1}) = 1.

Problem 11

Given cycles

\sigma = (1523)(47)

\tau = (183462)((57))

We compute

\sigma\tau = (18)(2543657)

And the orders of \sigma and \tau as the lcm of the cycle lengths.

Problem 12

Let G be a group with a,b,c \in G and a unique x such that axb = c.

We can prove x exists, because a^{-1},b^{-1} \in G, and thus

x = a^{-1}cb^{-1}.

Suppose there exists two solutions x_1,x_2.

Problem 13

Let A be an abelian group, and B be a subgroup of A. Then A/B is also abelian.

Let a_n \in A, b_n \in B, and n \in \mathbb{N}. Because were constructing a set A/B, we know that

aBa^{-1} \subseteq B

For any b \in B. Thus B can be though of as \ker(\phi) where

\phi \colon A \to A/B

Thus A/B is the group containing the fibers of \phi. The fibers of \phi will be denoted with x_n \in A/B.

Each x_n is related to a single a_n, so

a_1 \star a_2 = a_2 \star a_1 \implies x_1 \star x_2 = x_2 \star x_1

Problem 15

Prove the identity element e of a group is unique.

Towards a contradiction, let G be a group with two identity elements e_1, and e_2.

Problem 18

Let G be a group with identity e. Suppose that the order of G is equal to n. Prove x^n = e.