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{\bf Math 113 Homework 3, selected solutions}
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{\bf 1. Book exercise 3.3: is $\phi:(\Z, +)\to (\Z, +)$ given by $\phi(n) = 2n$ an isomorphism?} No. It is a homomorphism, since $\phi(a+b) = 2(a+b) = \phi(a) + \phi(b).$ It is also injective, since $2a = 2b$ implies $a = b$ for $a, b\in \zz$. But it is not surjective: $1$ is not $2a$ for any integer $a$!

BTW, it is a little surprising that there is a function from $\Z$ to $\Z$ that is injective but not surjective: wouldn't this imply that $\Z$ is ``bigger'' than itself? Turns out this is not the case for infinite sets: you can map them to themselves injectively with ``room left over''. If you're curious, you can look up ``infinite hotel'' online (or if you want to learn in a little more depth, look up ``cardinality'' on wikipedia).

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{\bf 3.} {\bf (a) Show that ``two-sided cancellation'' holds in groups. I.e., if $a, b, \tilde{b}, c$ are elements of a group $G$  with operation $\cdot$, and $a\cdot b\cdot c = a \cdot \tilde{b}\cdot c$ then $b = \tilde{b}$.}

Say $abc = a\tilde{b} c.$ Take both sides of the identity, multiply on the left by $a^{-1}$ (which exists since we're working in a group) and on the right by $c^{-1}.$ We obtain: $$b = a^{-1}\cdot a\cdot b\cdot c\cdot c^{-1} = a^{-1}\cdot a\cdot \tilde{b}\cdot c\cdot c^{-1} = \tilde{b},$$ so $b = \tilde{b},$ as desired. (Notice that when applying this argument it is important that I multiplied by $a^{-1}$ on the left and by $b^{-1}$ on the right -- if I was careless about order of factors, they would not cancel, since groups are not in general abelian!) 

\noindent {\bf (b) If $G$ is a group and $g\in G$ is a fixed element, show that the function $i_g$ with $i_g(x) = g\cdot x \cdot g^{-1}$ is an isomorphism of $G$ with itself.} Three things to show. First, homomorphism property:
$$i_g(x)\cdot i_g(y) = (g\cdot x \cdot g^{-1})(\cdot g \cdot y \cdot g).$$ By associativity, we can first multiply the $g^{-1}\cdot g$ in the middle \emph{(as this involves changing parenthesization, but not order of factors in the expression)}. Giving us: $$=g\cdot x \cdot e\cdot y\cdot g^{-1}= g\cdot x\cdot y\cdot g^{-1} = i_g(xy).$$ This concludes checking the homomorphism property. Now it remains to check bijectivity. As usual, this is done in two steps: \emph{Surjectivity}: the value $x$ is $i_g(g^{-1}xg)$ (since $\phi_g(g^{-1}xg) = g(g^{-1}xg)g^{-1} = e\cdot x \cdot e = x$.) And \emph{injectivity:} say $i_g(x) = i_g(\tilde{x}).$ Then $gxg^{-1} = g\tilde{x} g^{-1},$ and $x = \tilde{x}$ by part (a). 
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