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If you have a $\phi$-invariant, normal subgroup $N$ (so $\phi(N)=N$) of a finite group $G$, for an $\phi$, then you get an induced automorphism of $G/N$ by $gN\mapsto \phi(gN)=\phi(g)N$. The order of the induced automorphism is a divisor of $\phi$.

If $\phi$ is a regular automorphism (only fixed point is the identity), then the induced automorphism is regular as well. Is the order of the induced automorphism equal to the order of $\phi$ if it is regular?

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up vote 4 down vote accepted

No, not necessarily. Taking $N=G$, the induced automorphism is the only automorphism of the trivial group. This means it has order 1. The morphism $\phi$ need not have order 1.

Edit for an example with proper subgroups: Take two groups $G$ and $H$ and automorphisms $\phi:G\to G$ and $\varphi:H\to H$. Consider the automorphism $\phi\oplus\varphi$ of $G\oplus H$. It has order the least common multiple of the orders of $\phi$ and $\varphi$. Taking the normal subgroup to be $G$, the induced automorphism is essentially $\varphi$, which does not necessarily have the same order as $\phi\oplus\varphi$. (oh, and make sure $\phi\oplus\varphi$ is regular)

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That is a very good point. How about for proper subgroups? In particular, I'm working with the quotients of the lower central series. – Ske Apr 24 '12 at 9:04
It also doesn't hold for proper subgroups. Consider the cyclic group $C_9 = \langle c \rangle$ of order 9 and the automorphism $\phi: c \mapsto c^2$. If I'm not mistaken, $\phi$ is regular of order 6, but the induced automorphism on $C_9 / C_3$ is of order 2. – m_l Apr 24 '12 at 9:06
Thank you both. Very helpful. – Ske Apr 24 '12 at 9:21

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