How many $3$-Sylow subgroups are there in $\mathbb{Z}_{7}\rtimes_{\rho}\mathbb{Z}_{6}$ with $|\ker\rho| = 2$?

Question

Let $G=\mathbb{Z}_{7}\rtimes_{\rho}\mathbb{Z}_{6}$ with $|\ker\rho| = 2$. How many $3$-Sylow subgroups are there in $G$?

I know that the number is $1$ or $7$, but I'm stuck.

Answer 1

If the number of Sylow $3$-subgroups is $1$, then the Sylow $3$-subgroup is normal and trivially intersects the normal Sylow $7$-subgroup, whence it commutes with the Sylow $7$-subgroup element-wise. So it is contained in the kernel of $\rho$, a contradiction. So the number is $7$.

Answer 2

The question is compactly answered and accepted already, as this message is written, so consider it as a comment. We can in fact construct explicitly the group $G$ as a group of matrices over the field $F=\Bbb F_7$ with seven elements. Note first that the two groups $N=(\Bbb F_7,+)$ and $H=(\Bbb F_7^\times,\cdot)$ (obtained from $F$ by applying the one or the other forgetful functor) are isomorphich with the groups from the OP involved in the semidirect product. We have an action by multiplication of $H=(\Bbb F^\times,\cdot)$ on $N=(\Bbb F_7,+)$. Now we let each unit $u$ act on an $x$ by the rule: $$ \rho_u(x)=u^2x\ . $$ It is clear that only $\pm1$ are acting trivially, so the kernel of $\rho$ has two elements. The multiplication of two elements $(x,u)\in G$ and $(y,v)\in G$, $x,y\in N=\Bbb F_7$, $u,v\in H=\Bbb F^\times_7$, can be written $$ (x,u)(y,v)=(x+u^2y,uv)\ . $$ We have also a matrix representation for $(x,u)$ as the $3\times 3$ matrix over $\Bbb F_7$: $$ A(x,u)= \begin{bmatrix} 1 \\ \cdot & u\\ x &\cdot &u^2 \end{bmatrix} \ . $$ Now it is simple to exhibit all elements $(x,u)$ of order $3$ in this group, corresponding to matrices of the above shape of (multiplicative) order $3$. First of all, $u$ has order $1$ or $3$, so $u\in\{1,2,4\}$. The $1$ is easily excluded.

Then the computation of $$ (x,u)^3=(\ x(1+u^2+u^4)\ ,\ u^3\ ) $$ shows that each choice of $x$ is possible. So we have $2\times 7$ elements of order exactly three, concluding that there are $7$ ($3$-Sylow) groups with $3$ elements.

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Computer aid, here sage:

sage: F = GF(7)
sage: A = matrix(F, 3, 3, [1,0,0, 0,1,0, 1,0,1])
sage: B = matrix(F, 3, 3, [1,0,0, 0,3,0, 0,0,9])
sage: # 3 is a generator of (F*, .)
sage: G = MatrixGroup([A, B])
sage: G.order()
42
sage: G.structure_description()
'C2 x (C7 : C3)'
sage: P = G.as_permutation_group()

sage: for W in P.subgroups(): 
....:     if W.order() == 3: 
....:         print(f"Subgroup of order 3. Is it normal? {W.is_normal()}")
....:               
Subgroup of order 3. Is it normal? False
Subgroup of order 3. Is it normal? False
Subgroup of order 3. Is it normal? False
Subgroup of order 3. Is it normal? False
Subgroup of order 3. Is it normal? False
Subgroup of order 3. Is it normal? False
Subgroup of order 3. Is it normal? False
sage: