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# System prepended metadata

title: Mat 150 Homework 1

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# Mat 150 Homework 1

Problem 1: 
 The set of $2 \times 2$ matricies with entries from $\mathbb{R}$ forms a group under matrix addition. 
 
 proof: 
 Let A be the set of  $2 \times 2$ matricies with entries from $\mathbb{R}.$ A is a group under addition because it adheres to the group axioms. 
 
 $G1$: $\forall X,Y,Z \in A, (XY)Z = (X + Y) + Z = X + (Y + Z)= X(YZ)$. (Associativity for matrix addition)
 $G2$: $\exists 0_2\in A, \forall X \in A, X + 0_2 = X$ (Additive identity property for matrix addition)
 $G3: \forall X \in A, \exists Y \in A, X + Y = 0_2$. (Additive inverse property for matrix addition) 
 
 However; the set of $2 \times 2$ matricies does not forma  group under matrix multiplication since the set A does not satisfy the G3 axiom, since not all matricies in the set are nonsingular and thus do not have inverses. 
 
Problem 2: 
Consider that for $a,b, c \in G$, 
\begin{align*}
ab &= ac \\
a^{-1}ab &= a^{-1}ac\\
1b &= 1c &&\text{	(G3)}\\
b &= c &&\text{	(G2)}
\end{align*}

Problem 3 
Proof: 
(=>) We will first prove that if S is a group, then S has exactly 1 element. 
By $G2, \exists 1 \in G, \forall a \in G, 1a = a$. 
Suppose that S contains more than 1 element, and consider the element $b \in G$ such that $b \neq a$
By the definition of *, it is given that $aa = a * a = a$ and $ba = b * a = a$. 
However, by Corallary 4 which states that the identity is unique, it follows that $b = a = 1$. This contradicts the initial assumption that $b \neq a$. Thus; we conclude that S has only one element - the identity. 

(<=) We will now prove that if S has exactly 1 element, then it is a group. 
Let $S = {a}$. 
G1 is satisfied by default. 
G2 is true by definition of *, $aa = a*a = a$
G3 - 
By definition of *, $a^{-1}a = a^{-1} * a = a$. By corollary 4 it follows that $a^{-1} = a$, i.e that the a (the identity) is its own inverse - which satisfies G3. 




