Following this page, in the classification of groups of order $273$, the product of the Sylow group $S:=C_7C_{13}\simeq C_{7}\times C_{13}$ is normal, hence can be acted on by the Sylow $C_3$ to produce semidirect products.

Since $\operatorname{Aut}(S)\simeq C_6\times C_{12}$, one only needs to find an order 3 element there to produce the homomorphism $C_3\rightarrow \operatorname{Aut}(S)$.

In particular, if $C_6=\langle x\rangle$ and $C_{12}=\langle y\rangle$, up to choice of generators of $C_3$, the group is classified by the 5 actions corresponding to: $(1,1), (x^2,1), (1,y^4), (x^2,y^4), (x^{-2},y^4)$.

It is easy to see that the center of the corresponding groups has order $273, 13, 7, 1, 1$ respectively. So my question is: how do we show that the last two groups are actually non-isomorphic?

P.S. In a similar example with $(C_7\times C_7)\rtimes C_3$ on the same page, the last two groups can be distinguished since every subgroup of order $7$ can be shown to be normal in one but not the other.


Edit: I didn't realize you were talking about $x$ and $y$ being in the automorphism group, and not the group itself. I've edited below to correct this.

We can take $\alpha$ as a generator of $C_7$, and then $x(\alpha)=\alpha^5$. So $x^2(\alpha) = \alpha^4$ and $x^{-2}(\alpha) = \alpha^2$.

We can take $\beta$ as a generator of $C_{13}$, and then $y(\beta)=\beta^2$, and so $y^4(\beta) = \beta^3$.

Consider your groups as $G=C_{91}\rtimes C_3$ and $H=C_{91}\rtimes C_3$. We can write $z$ as the generator of $C_{91}$, and $g$ and $h$ as the generator of the $C_3$ subgroups.

Then in $G$, we have $g^{-1}zg=z^{-10}$, since $-10$ is $4\pmod{7}$ and $3\pmod{13}$.

In $H$, we have $h^{-1}zh=z^{16}$, since $16$ is $2\pmod{7}$ and $3\pmod{13}$.

If $f:G\rightarrow H$ was an isomorphism, then we would have $f(z)=z^k$ for some integer $k$. We would also have $f(g)$ is some conjugate of $h$ or $h^2$, which thus acts the same as $h$ or $h^2$ on $z$.

So then

\begin{align*} z^{-10k} &= f(z^{-10})\\ &= f(g^{-1}zg)\\ &= f(g^{-1})z^kf(g)\\ &= z^{16k}\text{ or }z^{74k} \end{align*}

This implies $z^{26k}=1$ or $z^{84k}=1$, so that $7$ or $13$ divides $k$ respectively, which means $z^k$ is not a generator of $C_{91}$ in $H$. This is the contradiction.

This same idea can be used whenever you have a semidirect product $A\rtimes B$, and $A$ is abelian and $B$ is cyclic. It can be used to differentiate different semidirect products coming from non-conjugate images of $B$ in $\textrm{Aut}(A)$.

  • $\begingroup$ This is satisfying~ By the way I think $f(g)$ is some conjugate of $h$ or $h^2$? $\endgroup$
    – Kamineko
    Sep 12 '18 at 15:24
  • $\begingroup$ Also $z^{26k}=1$ already says $z^k$ has order 26 hence not a generator. $\endgroup$
    – Kamineko
    Sep 12 '18 at 15:24
  • $\begingroup$ @Kamineko: yes, I had blindly thought that $h$ and $h^2$ would be conjugate in $H$, but they are not, I'll edit. $\endgroup$
    – Steve D
    Sep 12 '18 at 15:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.