Groups of the form $(\Bbb Z_{13} \times \Bbb Z_7)\rtimes \Bbb Z_3$

Question

Consider the semi-direct product:

$(\Bbb Z_{13} \times \Bbb Z_7)\rtimes \Bbb Z_3$

To construct a group $G$, we need homomorphisms $\theta$: $\Bbb Z_3 \rightarrow \text{Aut}(\Bbb Z_7)$ and $\theta_2$: $\Bbb Z_3 \rightarrow \text{Aut}(\Bbb Z_{13})$.

Consider the set of coordinates

$(x,y,z)$

with $x\in \Bbb Z_{13}$, $y\in \Bbb Z_7$, and $z\in \Bbb Z_3$.

From the homomorphism, we can define a group law

$(x,y,z) * (x_2,y_2,z_2) = (x+\theta_2(z)(x_2), y+\theta(z)(y_2), z+z_2)$.

More specifically, consider the groups given by their operations:

$G_1 = (x,y,z) * (x_2,y_2,z_2) = (x+x_23^{z}, y+y_21^{z}, z+z_2)$.

$G_2 = (x,y,z) * (x_2,y_2,z_2) = (x+x_23^{z}, y+y_22^{z}, z+z_2)$.

$G_3 = (x,y,z) * (x_2,y_2,z_2) = (x+x_23^{z}, y+y_24^{z}, z+z_2)$.

It is clear why $G_1$ is not isomorphic to $G_2$ or $G_3$:

$G_1$ has $156$ elements of order $21$, whereas the other two groups have no such elements of order $21$.

I am struggling to see why $G_2$ and $G_3$ are non-isomorphic? If two groups have the same number of elements of any given order, doesn't that imply they are isomorphic?

Both $G_2$ and $G_3$ have $182$ elements of order $3$, $6$ of order $7$, $12$ of order $13$, and $72$ of order $91$.

Answer 1

They are not isomorphic. In fact they are $\mathtt{SmallGroup}(273,3)$ and $\mathtt{SmallGroup}(273,4)$ in the small groups databases in GAP and Magma. This example shows that it is possible for two non-isomorphic finite groups to have the same numbers of elements of each order.

In many examples of this type, there is no "nice" proof of non-isomophism resulting from the two groups having different invariants, and you just have to do it using a direct argument.

An isomorphism would have to map the unique Sylow 7- and 13- subgroups of one group to the other and, since all Sylow 3-subgroups are conjugate, you could assume that it maps a generator $z_1$ of $C_3$ in the first group to either $z_2$ or $z_2^{-1}$ in the second group. In either case, you will find that there are no isomorphisms between the Sylow 7- and 13- subgroups that give rise to an isomorphism between the groups.

Answer 2

I expected this to have been covered on the site already, but 20 minutes of searching didn't help, so here comes an elementary explanation.

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Both groups $G_2$ and $G_3$ have the form $C_{91}\rtimes C_3$. If here $C_{91}=\langle a\rangle$ and $C_3=\langle b\rangle$ then in $G_2$ we have $$ bab^{-1}=a^{16}\qquad(*) $$ as $16\equiv3\pmod{13}$ and $16\equiv2\pmod 7$. In other words $G_2=C_{91}\rtimes_{\phi_2}C_3$ with $\phi_2:C_3\to Aut(C_{91})=\Bbb{Z}_{91}^*$ determined by $\phi_2(b)=16$.

On the other hand, in $G_3$ we have $$ bab^{-1}=a^{81}\qquad(**) $$ as $81\equiv3\pmod{13}$ and $81\equiv4\pmod7$. So $G_2=C_{91}\rtimes_{\phi_3}C_3$ with $\phi_3:C_3\to Aut(C_{91})=\Bbb{Z}_{91}^*$ determined by $\phi_2(b)=81$.

Finding the exponents $16$ and $81$ were instances of the Chinese Remainder Theorem.

The groups $H_2=\langle 16\rangle=\{1,16,74\}\le\Bbb{Z}_{91}^*$ and $H_3=\langle 81\rangle=\{1,81,9\}$ intersect trivially, and this suffices:

The relation $(*)$ holds with any element of order $91$ in place of $a$. Similarly $(**)$ holds for every element of order $91$ in place of $a$. In both $G_2$ as well as $G_3$ all the elements $b'$ of order three are conjugates with either $b$ or $b^2$. Furthermore, we only need powers of $a$ to get the conjugacy.

It follows that any element $A\in G_2$ of order $91$ and any element $B$ of order $3$ satisfy $BAB^{-1}=A^k$ with $k\in H_2\setminus\{1\}$. A similar relation holds in $G_3$, but this time with $k\in H_3\setminus\{1\}$. Therefore no isomorphism between $G_2$ and $G_3$ exists.

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Leaving it to others to formulate a suitable more general result about non-isomorphism of semidirect product of two cyclic groups of coprime orders.