Let $A(d)$ be the following assertion : If $H$ is a group of order $102$, then $H$ has a normal subgroup of order $d$. Prove as many statements $A(d)$ as possible.
Here's what I've come up with so far:
$A(102)$ and $A(1)$ follow trivially, since $H$ is normal in $H$ and the trivial subgroup $\{e\}$ is normal in $H$.
By the Sylow Theorems, since $102 = 2 \cdot 3 \cdot 17$, denoting $n_k$ by the number of $k$-Sylow subgroups of $H$, we get $n_2 | 51$ and $n_2 \equiv 1$ (mod $2$), $n_3 | 34$ and $n_3 \equiv 1 $(mod $3$), and $n_{17} | 6$ and $n_{17} \equiv 1$ (mod $17$) $\Rightarrow$ $n_2 = 1,3,17,51$ , $n_3 = 1, 34$ , $n_{17} = 1$.
Thus, by the above, we have a normal Sylow $17$-subgroup of $H$ $\Rightarrow$ $A(17)$ is proven.
Therefore, I've proven $A(1), A(102)$, and $A(17)$ so far.
Are there other assertions $A(d)$ that can be proven here ? I don't see how to prove more assertions $A(d)$ than the three above from Sylow Theory. I'm also not sure of other techniques and machinery to detect what normal subgroups $H$ must necessarily have.
Thanks!
Best Answer
If a group has twice odd order, then it has a normal subgroup of index 2. See for example A Group Having a Cyclic Sylow 2-Subgroup Has a Normal Subgroup.
So $H$ has a normal subgroup $N$ of order $3\cdot 17$. Now, show that Sylow subgroups of $N$ are normal. Then they are characteristic in $N$, so normal in $H$. That will give you $A(3)$, $A(17)$ and $A(3\cdot 17)$. It turns out that this is all you can prove. To see this, consider the dihedral group of order $102$. You will see that it has no proper normal subgroup of even order. (Just find the group generated by the conjugacy class of an element of order $2$. This part of the argument works for every group of twice odd order.)