Suppose $|G|=154$ and $G$ is not decomposable in the sense that there are no proper subgroups $H$ and $K$ of $G$ such that $G \cong H \times K$. Show that $G \cong D_{154}$.
Here is my attempt: Let $P \in \mathrm{Syl}_{7}(G)$ and $Q \in \mathrm{Syl}_{11}(G)$. Using Sylow theory we can show that that at least one $P$ or $Q$ is normal in $G$, which implies that $N=PQ \le G$. In fact $[G\colon N]=2$ so $N \trianglelefteq G$.
Let $H \in \mathrm{Syl}_{2}(G)$. We have $HN \le G$ and $|HN|=|H||N|=|G|$ because $|H \cap N|=1$ (since the only element with order dividing both $77, 2$ is $e$). Hence $G = HN$.
To see that $G \cong N \rtimes H$, we need $\varphi \in \mathrm{Hom}(H, \mathrm{Aut}(N))$ that is non-trivial. We have \begin{align*} \varphi\colon H &\to \mathrm{Aut}(N)\\ h &\mapsto \varphi_{h}\colon n \mapsto hnh^{-1}; \end{align*} If $h \neq e$ then $|h|=2$. We always have $|\varphi_{h}| \mid |h|$. To form a non-trivial semi-direct product we need a $\varphi_{h} \in \mathrm{Aut}(H)$ with order $2$.
The proof would be done if I could prove that we must have $\varphi_{h}\colon n \mapsto hnh^{-1}=n^{-1}$ and then $G \cong \langle n, h \mid n^{77}=h^{2}=e, hnh^{-1}=n^{-1} \rangle = D_{154}$.
Question: How can I show that inversion is the only possible automorphism of order $2$?
In fact, both $P$ and $Q$ are normal, because by Sylow theorems, $n_{11} \mid 14$ and $11\mid (n_{11}-1)$. So we get that $n_{11}=1$. Same discussion for $n_{7}$. Now by normalizer-centralizer (or any otherway), we see that $|G:C(Q)|= 1$ or $2$. If $|G:C(Q)|= 1$, $G\cong Q\times K$ (we know that such $K$ exists), a contradiction. So $|G:C(Q)|= 2$. Note that since $G\cong (Q\times P)\rtimes H$, we just need to prove that $H$ acts Frobeniusly on $P$, since we know that it acts Frobeniusly on $Q$. Again $|G:C(P)|= 1$ or $2$, and if $|G:C(P)|= 1$, $G\cong P\times K$ (we know that such $K$ exists), a contradiction. Hence $|G:C(P)|= 2$, which means $H$ acts Frobeniusly on $P$. Now we conclude that $H$ acts Freobeniusly on $N$, and that means $G\cong D_{154}$.