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Embed Dark Mode Prev Up Next \(\require{mathtools}\require{color} \setcounter{MaxMatrixCols}{15}
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Section Isomorphisms of Groups
Worksheet Part 1: Isomorphisms
The goal of these problems is to practice working with isomorphisms of groups.
1. Let
\(H\) be the subgroup of rotations in
\(D_4\text{.}\) Show that
\(H\cong \Z_4\text{.}\)
2. Show that
\(U(8)\) is not isomorphic to
\(U(10)\text{.}\)
Solution .
As in
\(U(10) \not\isom U(12)\) it is not possible for a map from
\(U(10)\) to
\(U(8)\) to be both operation preserving and bijective. Concretely, we must have
\begin{equation*}
\phi(9)=\phi(3\cdot 3)=\phi(3)\cdot \phi(3)=(\phi(3))^2=1\text{,}
\end{equation*}
since all elements of \(U(8)\) square to 1. But then \(\phi\) is not injective since \(\phi(1)=1\) as well.
Definition 57 . Automorphism of a Group.
An
automorphism of a group
\(G\) is an isomorphism from
\(G\) to itself. The set of all automorphisms of a group
\(G\) forms a group under composition, called the
automorphism group of
\(G\text{,}\) denoted
\(\Aut(G)\text{.}\)
3. Let
\(\R^+\) be the group of positive real numbers under multiplication. Show that the mapping
\(\phi(x)=\sqrt{x}\) is an automorphism of
\(\R^+\text{.}\)
Solution .
We have
\begin{equation*}
\phi(xy)=\sqrt{xy}=\sqrt{x}\sqrt{y}=\phi(x)\phi(y)
\end{equation*}
for any \(x,y\in \R^+\text{,}\) so \(\phi\) is operation preserving. Also, \(\phi\) is one-to-one since \(\phi(x)=\phi(y)\) implies \(\sqrt{x}=\sqrt{y}\) and hence \(x=y\text{.}\) Finally, \(\phi\) is onto since for any \(z\in \R^+\text{,}\) we have \(\phi(z^2)=z\text{.}\)
4. Conjugation is an Automorphism.
Let
\(G\) be any group and
\(a\in G\text{.}\) Show that the map
\(\phi_a(g)=aga^{-1}\) is an automorphism of
\(G\text{.}\)
Solution .
We have
\begin{equation*}
\phi_a(gh)=a(gh)a^{-1}=(aga^{-1})(aha^{-1})=\phi_a(g)\phi_a(h)
\end{equation*}
for any \(g,h\in G\text{,}\) so \(\phi_a\) is operation preserving. Also, \(\phi_{a^{-1}}\) is an inverse map for \(\phi_a\text{,}\) since
\begin{equation*}
\phi_{a^{-1}} (\phi_a(g))=a^{-1}(aga^{-1})a=g=a(a^{-1}ga)a^{-1}=\phi_a(\phi_{a^{-1}}(g))
\end{equation*}
so \(\phi_a\) is bijective and hence an automorphism.
