Inner Automorphisms

The goal of these problems is to continue working with isomorphisms of groups and investigate the automorphism group and inner automorphism group of a group.

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Definition 64. Inner Automorphism.

If \(G\) is a group and \(a\in G\text{,}\) the inner automorphism of \(G\) induced by \(a\) is the function \(\phi_a\) defined by

\begin{equation*} \phi_a(x)=axa^{-1}\text{.} \end{equation*}

The set of all inner automorphisms of a group \(G\) forms a subgroup of \(\Aut(G)\text{,}\) called the inner automorphism group of \(G\text{,}\) denoted \(\Inn(G)\text{.}\)

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1.

You showed on the previous sheet that this is indeed an automorphism of \(G\text{.}\) Now prove that \(\Inn(G)=\{ \phi_a \mid a \in G \}\) is a group under function composition.

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Solution.

These are a subset of the automorphisms of \(G\text{,}\) so we use the Two-Step Subgroup Test. Since \(G\) is nonempty, so is \(\Inn(G)\text{.}\) Let \(\phi_a, \phi_b \in \Inn(G)\text{.}\) Then we have

\begin{equation*} \phi_{a} (\phi_b(x)) = a(bxb^{-1})a^{-1} = (ab)x(ab)^{-1}=\phi_{ab}(x)\text{,} \end{equation*}

so \(\phi_a \circ \phi_b = \phi_{ab}\) is an inner automorphism. Applying the same fact with \(b=a^{-1}\) shows that \(\phi_a^{-1}=\phi_{a^{-1}}\) is an inner automorphism. Thus \(\Inn(G)\) is a subgroup of \(\Aut(G)\text{.}\)

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2.

Compute the action of the inner automorphism \(\phi_{3}\) on \(\Z_8\text{.}\)

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Solution.

We have

\begin{equation*} \phi_3(x)=3+x-3=x \end{equation*}

for all \(x\in \Z_8\text{,}\) so \(\phi_3\) is the identity map on \(\Z_8\text{.}\) This holds more generally: if \(G\) is an Abelian group, then \(\Inn(G)\isom\{e\}\text{.}\)

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3.

Compute the action of the inner automorphism \(\phi_{(123)}\) on \(S_3\isom D_3\text{.}\) Draw three Cayley diagrams of \(S_3\) to illustrate this action: one the Cayley diagram of the original group with generators \((123)\) and \((12)\text{,}\) one the “relabeling” by \(\phi_{(123)}\) of the first diagram, and one the “rewiring” by \(\phi_{(123)}\) of the first diagram.

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Solution.

It suffices to compute the images of a generating set of \(S_3\text{,}\) e.g. \((123)\) and \((12)\text{.}\) We have

\begin{equation*} \phi_{(123)}((123))=(123) \quad \phi_{(123)}((12))=(23)\text{.} \end{equation*}

So we get the three Cayley diagrams below.

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Cayley diagram of S_3 with generators (123), (12) — Figure 65. Cayley diagrams for \(S_3\) with generators \((123), (12)\)
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Relabeled diagram fixing the rotations and rotating the reflections clockwise — Figure 66. Relabeling of the Cayley diagram under the action of \(\phi_{(123)}\)
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Rewired diagram fixing the actions of the red arrows and rotating the actions of the blue arrows counterclockwise — Figure 67. Rewiring of the Cayley diagram under the action of \(\phi_{(123)}\)
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4.

While \(\Inn(\Z_{8})=\{e\}\) is trivial, the full automorphism group of \(\Z_8\) is not. One such automorphism is the map \(\phi(x)=3x \pmod 8\text{.}\) Draw three Cayley diagrams of \(\Z_8\) to illustrate the action of this automorphism: one the Cayley diagram of the original group with generator \(1\text{,}\) one the “relabeling” by \(\phi\) of the first diagram, and one the “rewiring” by \(\phi\) of the first diagram.

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Solution.

It suffices to compute the images of a generating set of \(\Z_8\text{,}\) e.g. \(1\text{.}\) We have

\begin{equation*} \phi(1)=3\text{.} \end{equation*}

So we get the three Cayley diagrams below.

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