\magnification=\magstep1
\centerline{Mathematics 243, section 1 -- Algebraic Structures}
\centerline{Selected Solutions -- Problem Set 7}
\centerline{November 20, 2003}
\bigskip
\noindent
3.4/1.  Let $\phi : G \to H$ be a group isomorphism.   Then we 
know $\phi$ is 1-1, onto and satisfies the group homomorphism
property $\phi(xy) = \phi(x)\phi(y)$
for all $x,y \in G$.  Since $\phi$ is 1-1 and onto, there exists
an {\it inverse mapping} $\phi^{-1} : H \to G$.  Since $\phi = 
(\phi^{-1})^{-1}$, $\phi^{-1}$ is also one-to-one and onto.
Hence we only need to show that $\phi^{-1}$ is also a 
group homomorphism.  Let $w, z\in H$.  Since $\phi$ is onto,
there exist $x,y\in G$ such that $\phi(x) = w$ and $\phi(y) = z$
(so $\phi^{-1}(w) = x$ and $\phi^{-1}(z) = y$.  Hence
$$wz = \phi(x)\phi(y) = \phi(xy) \Rightarrow \phi^{-1}(wz) = xy 
= \phi^{-1}(w)\phi^{-1}(z)$$
This is what we wanted to show.
\bigskip
\noindent
3.4/2.  Although the notation $\phi_2\phi_1$ is slightly ambiguous
in the problem, the only consistent interpretation from the context
is that this is the {\it composite mapping}:
$$\phi_2 \circ \phi : G_1 \to G_3$$
From work in Chapter 1, we know that the composition of 1-1 mappings
is 1-1 and the composition of onto mappings is onto.  Hence
since $\phi_1$ and $\phi_2$ are group isomorphisms, $\phi_2\circ \phi_1$
is also 1-1 and onto.  Then, if $x,y\in G_1$ are arbitrary, we have
$$(\phi_2\circ \phi_1)(xy) = \phi_2(\phi_1(xy)) = \phi_2(\phi_1(x)\phi_1(y))
= \phi_2(\phi_1(x))\phi_2(\phi_1(y))$$
since both $\phi_1$ and $\phi_2$ have the group homomorphism property.
It follows that $\phi_2\circ\phi_1$ is also an isomorphism of groups.
\bigskip
\noindent
3.4/21.  Let $G$ and $H$ be any two cyclic groups of order $n$.
By definition, this means that $G = \langle a\rangle$, 
$H = \langle b\rangle$ for some $a\in G$ and $b \in H$, and 
moreover, $a^n = b^n = e$ and $n$ is the smallest strictly 
positive integer with this property.  Therefore,
$$\eqalign{
G &= \{e = a^0, a = a^1, \ldots, a^{n-1}\}\cr
H &= \{e = b^0, b = b^1, \ldots, b^{n-1}\}\cr}$$
Define $\phi: G \to H$ by $\phi(a^k) = b^k$ for all $k \in {\bf Z}$.
Then $\phi$ is clearly one-to-one and onto since it maps distinct
elements of $G$ to distinct elements of $H$ and every element of $H$
is the image of some element of $G$.  Finally, letting $x = a^\ell$
$y = a^m$ be any two elements of $G$, we have
$$\phi(a^\ell \cdot a^m) = \phi(a^{\ell + m}) = b^{\ell + m} = 
b^{\ell} \cdot b^m = \phi(a^\ell)\phi(a^m)$$
so $\phi$ is an isomorphism of groups.
\bigskip
\noindent
3.5/10.  Let $H$ be a homomorphic image of $G$ (that is, there exists
an onto group homomorphism $\phi : G \to H$).  Let $w,z$ be arbitrary
elements of $H$.  Then since $\phi$ is onto, $w = \phi(x)$ and 
$z = \phi(y)$ for some $x,y$ in $G$.  We have, using commutativity in 
$G$:
$$wz = \phi(x)\phi(y) = \phi(xy) = \phi(yx) = \phi(y)\phi(x) = zw$$
Hence $H$ is also abelian.
\bye
