In the hope that this will be the definitive answer to understanding groups of order $105$, I will talk about the ways to solve this.
The question assumes that the Sylow $3$-subgroup is normal in $G$. The condition on the Sylow $3$-subgroup here is necessary. There are two groups of order $105$, both with a normal Sylow $5$- and $7$-subgroup, but one is cyclic and the other is $C_5\times F_{21}$, where $F_{21}$ is a non-abelian group, the normalizer of a Sylow $7$-subgroup of $A_7$.
The fastest way to proceed is to notice that $P_3$, the Sylow $3$-subgroup, is not only normal but central. To see this, you can recall that $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\mathrm{Aut}(H)$, which has order $2$ in this case. The from scratch method is to notice that $C_3$ has only two non-identity elements, so for any element $g\in G$, $g^2$ must act trivially on $P_3$. But $|G|$ is odd, so every element is a square, and $P_3$ is central.
At this point, there are two ways to proceed. The first is to notice that $G/P_3$ has order $35=5\times 7$, and groups of order $35$ are cyclic. If $G/Z(G)$ is cyclic then $G$ is abelian, and we are done. (Clearly $G$ is therefore in fact cyclic.)
The alternative proof is to note that $P_3\leq C_G(P_5)$ and $P_3\leq C_G(P_7)$. Thus $|C_G(P_5)|\geq 15$, and $|C_G(P_7)|\geq 21$. (Recall that $C_G(P_q)\leq N_G(P_q)$ and $n_q$, the number of Sylow $q$-subgroups, is equal to $|G:N_G(P_q)|$.) From Sylow's theorem ($n_q\equiv 1\bmod q$) we see that $n_5=n_7=1$, as needed.
If you don't want to do this you can count elements, although it's more subtle than most such arguments. Let's do this without the assumption that $n_3=1$, to obtain the full classification.
The number $n_5$ of Sylow $5$-subgroups is either $1$ or $21=3\times 7$. We want to prove the former, so assume the latter. Then there are $21\times 4=82$ elements of order $5$, and since $C_G(P_5)=P_5$, there are no elements of order $5n$ for any $n>1$. This leaves exactly $105-82=23$ elements of order not $5$, and these must have order $1$, $3$, $7$ or $21$. If $n_7\neq 1$ then $n_7=15$, but this is impossible as there are only $23$ elements left. So $n_7=1$, removing six elements of order $7$. There are seventeen elements left, so $n_3\leq 8$ (as each Sylow $3$-subgroup requires two elements of order $3$). Thus $n_3=1$ or $n_3=7$. If $n_3=7$ then that removes fourteen elements of order $3$, and the identity, so there are two elements remaining, which must have order $21$. But in any cyclic group of order $21$ there are twelve elements of order $21$, which is too many.
Thus $n_3=1$, and the Sylow $3$- and $7$-subgroups are both normal. Thus $P_3P_7$ is normal in $G$, has index $5$, and therefore contains every element of order dividing $21$. So where are the two remaining elements? This yields a contradiction, so $n_5=1$.
If $n_7\neq 1$ then $n_7=15$, as it must be $1$ modulo $7$. Again, you can obtain a contradiction as before, because $C_G(P_5)$ contains $P_7$ but $C_G(P_7)$ does not contain $P_5$. Let's try to count elements, and see what goes wrong. This yields $15\times 6=90$ elements of order $7$. There are five elements in $P_5$, leaving ten elements. Thus $n_3\leq 5$, so $n_3=1$. Thus we have a subgroup $P_3P_5$ of order $15$. This contains ten more elements (as we have already counted $P_5$), and so we have exactly the right number of elements, $105$.
If $15$ were a prime, then this would be fine. Then $7\mid (15-1)$ and there would be a map from $C_7$ into $\mathrm{Aut}(C_{15})$, which would have order $14$. But $15$ is not a prime, so we can obtain a contradiction using centralizers, as above, but element counting will not work in this case. The group $P_3P_5$ has normal subgroups $P_3$ and $P_5$, on which $P_7$ cannot act. Thus $P_3P_5$ is actually central, and $G/(P_3P_5)$ is cyclic, so $G$ is abelian. Alternatively, $P_3$ is central, so $P_3$ centralizes $P_7$. But $n_7=15$, so $P_7$ does not centralize $P_3$. This is a clear contradiction.
Thus $n_7=1$ as well. The subgroup $P_5P_7$ is a normal, cyclic, subgroup of order $35$. Since there is no map from $P_3$ to $\mathrm{Aut}(P_5)$, this is actually central. The subgroup $P_7P_3$, of order $21$, complements this, so $G\cong P_5\times P_7P_3$. If $n_3=1$, equivalently $P_3$ centralizes $P_7$, then you end up with an abelian (cyclic) group of order $21$. If $n_3=7$, equivalently $P_3$ acts non-trivially on $P_7$, then $P_3P_7$ is a Frobenius group of order $21$. This is the normalizer in $A_7$ of a Sylow $7$-subgroup.
$H$ is a subgroup of $G$ which is the direct product of $C_5$ and $C_3$. Both $C_5$ and $C_3$ have an automorphism of order $2$, which acts as $\phi(x)=x^{-1}$.
Note that $G/N$ has order $2$, so either $C_G(N)=N$ or $C_G(N)=G$. In the latter case, you get a direct product $C_5 \times C_3 \times C_2 = C_{30}$. So assume that $C_G(N)=N$. Now you can define a map $\phi: C_2 \to \operatorname{Out}(H)$ as the map defined by the action of $G/N$ by conjugation on $N$ (this is the NC theorem). There are three possibilities for $G/N = C_2$:
- It acts trivially on $C_5$, but not on $C_3$. In this case you get $C_5 \times (C_3 \rtimes C_2) = C_5 \times D_6$.
- It acts trivially on $C_3$, but not on $C_5$. In this case you get $C_3 \times (C_5 \rtimes C_2) = C_5 \times D_{10}$.
- It acts non-trivially on both $C_5$ and $C_3$. In this case, you get $(C_5 \times C_3) \rtimes C_2 = C_{15}\rtimes C_2 = D_{30}$.
Essentially, you are looking at how a $C_2$ can act on two factors, each of which has a single automorphism of order two (so the action is either that automorphism, or trivial).
For the second one, it's the same logic, but now you are looking at maps $\phi: C_2 \times C_2 \to C_2$ (where the latter is $\operatorname{Aut}(C_3)$). Let $C_2 \times C_2 = \langle a \rangle \times \langle b \rangle$. Take the preimage of $C_2 = \operatorname{Aut}(C_3)$: it can be one of three subgroups $\langle a \rangle$, $\langle b \rangle$, $\langle ab \rangle$. Call it $T$. Whatever it is, you can still write:
$$ G = (C_3 \rtimes T) \times C_2$$
as follows
- $ G = (C_3 \rtimes \langle a \rangle) \times \langle b \rangle$
- $ G = (C_3 \rtimes \langle b \rangle) \times \langle a \rangle$
- $ G = (C_3 \rtimes \langle ab \rangle) \times \langle a \rangle$
and so all the semidirect products are isomorphic.
Best Answer
Consider the special case where $\phi_2$ is constant equal to the identity, so that one of the semi-direct products is actually a direct product. The converse of your proposition in that case would be :
If $N \rtimes_{\phi_1} H \cong N\times H$ then there is some $g\in \operatorname{Aut}(N)$ such that $\phi_2(h)=g\phi_1(h)g^{-1}$ for all $h\in H$, which in turn implies that $\phi_1$ is constant equal to the identity as well.
But it is well-known that non-trivial actions can lead to direct products.