The semidirect product $(C_7\times C_{13})\rtimes C_3$

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 $\textrm{Aut}(A)$.

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