I have the following question and have become a bit stuck. Any help is greatly appreciated!

Let $Z=\langle z\rangle$ be the cyclic group of order $7$ (so for every $i=1,\dots,6$ there's an automorphism $f_i$ taking $z$ to $z^i$), so $Y=Aut(Z)= \{f_i\mid i=1,\dots,6\}$. Also, let $X=\langle x\rangle$ be a cyclic group of order $3$.

How do I compute the homomorphism $\phi:X\rightarrow Y=Aut(Z)$ sending $x$ to $f_2$?


This is extremely simple. $X=\{1,x,x^2\}$, $\phi(x)=f_2$. Thus $\phi(x^2)=\phi(x)\circ \phi(x)= f_2\circ f_2$. What is $f_2\circ f_2$?

  • $\begingroup$ Thank you so much, that's what I thought it was but I didn't think it would be as straightforward. So $f_2\circ f_2 = (x^2)^2 = x^4 = x$? $\endgroup$ – AKCJ Mar 18 '17 at 15:51
  • $\begingroup$ @AKCJ $f_2\circ f_2=f_4$. $x$ is an element of the domain. It is not $f_i$ for any $i$. $\endgroup$ – Matt Samuel Mar 18 '17 at 15:58
  • $\begingroup$ thanks! So if i'm doing calculations based on the semi-direct product of G : = X x Z, will |G|= |X||Z| = 3(7) = 21? $\endgroup$ – AKCJ Mar 18 '17 at 16:03
  • $\begingroup$ @AKCJ Yes. The underlying set of the semidirect product is the Cartesian product. $\endgroup$ – Matt Samuel Mar 18 '17 at 16:04
  • $\begingroup$ Thank you so much for your help! $\endgroup$ – AKCJ Mar 18 '17 at 16:06

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