Prove a group of order $48$ must have a normal subgroup of order $8$ or $16$.

Solution: The number of Sylow $2$-subgroups is $1$ or $3$. In the first case, there is one Sylow $2-$subgroup of order $16$, so it is normal.

In the second case, there are $3$ Sylow $2$-subgroups of order $16$. let $G$ act by conjugation on the Sylow $2$-subgroups. This produces a homomorphism from $G$ into $S_3$. The image cannot consist of just $2$ elements. Also, since no Sylow $2$-subgroup is normal, the kernel cannot have $16$ elements. The only possibility is that the homomorphism maps $G$ onto $S_3$, and so the kernel is a normal subgroup of order $48 / 6 = 8$.

My first question is the first sentence of the bolded part: "The image cannot consist of just $2$ elements." Why can't it have $2$ elements? The homomorphism will map $24$ elements of $G$ to the identity permutation of $S_3$, and the homomorphism will map $24$ elements of $G$ to the a $2-$cycle permutation of $S_3$. I don't see any contradiction here. If mapping $24$ elements of $G$ to a $2-$cycle permutation of $S_3$ implies that there is a normal subgroup of order $8$ or $16$, then I don't see why this implication is true.

My second question is about: "since no Sylow $2$-subgroup is normal, the kernel cannot have $16$ elements." I don't understand this statement either. I don't see any relationship at all between a Sylow $2$-subgroup not being normal and the kernel of the homomorphism. If the kernel of a homomorphism has $16$ elements, then I agree that these $16$ elements constitute a normal subgroup of $G$, but I don't see what that has anything to do with a Sylow $2$-subgroup being normal or not.

  • $\begingroup$ Isn't that the same as your previous question? They assumed in addition that the three Sylow 2-subgroups are not normal (If one of them is normal, you are done) Using this extra assumption, the image cannot have only 2 elements. Because if it is so, then the image has to be generated by an element of order 2, which is either $(12), (13), (23)$. Then (e.g.) if it is $(12)$, then $H_3$ (the third Sylow 2-subgroup) will be normal. $\endgroup$ – user99914 Oct 16 '15 at 4:39
  • $\begingroup$ On the other hand, it cannot have 3 elements, or the kernal $G \to S_3$ will have 16 element, that is a Sylow 2-subgroup. But then this subgroup will be normal and that violate your extra assumption. $\endgroup$ – user99914 Oct 16 '15 at 4:40
  • $\begingroup$ @JohnMa Yes it's the same question but I still don't understand it. I don't understand why if the image has $(12)$ then $H_3$ will be normal. Also why is the kernel of $16$ elements a Sylow $2-$subgroup and how do you know it's not a normal subgroup of order $16$ that's not a Sylow $2-$subgroup? $\endgroup$ – mr eyeglasses Oct 16 '15 at 4:45
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    $\begingroup$ No, there are 24 elements in $G$ that map $H_1 \to H_2$, so $gH_ig^{-1} = H_i$ is not satisfied for all $g\in G$. $\endgroup$ – user99914 Oct 16 '15 at 4:55
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    $\begingroup$ @JohnMa Oh wow, I think I realize now, that with our mapping (like the one I drew in my picture), $\textit{all}$ $48$ elements in $G$ are mapping $H_3 \to H_3$, because $24$ of them are mapping $H_3 \to H_3$ in the identity, and the other $24$ different elements are mapping $H_3 \to H_3$ in the $2-$cycle permutation $\endgroup$ – mr eyeglasses Oct 16 '15 at 4:59

You can try this way:


Now Number of Sylow $2$ subgroups of $G$ =$1+2k$ .Either $1+2k=1 or 3$ .If it is $1$ we are done.If it is $3$ then we have $3$ Sylow $2$ subgroups of $G$ of order $16$ say $H_1,H_2,H_3$ .then

$o(H_1H_2)=\dfrac{o(H_1)o(H_2)}{o(H_1\cap H_2)}$.Possible orders of $o(H_1\cap H_2)=2,4,8$ by lagrange's theorem/.

The only one which works here is $8$ (verify).

Now use the result that

For any $p-$ group $G$ having order $p^n$ ,if $H$ is a subgroup of order $p^{n-1}$ then $H$ is normal in $G$.(Try it;do inform me for any hints)


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