# Question about proof that group of order 48 must have a normal subgroup of order 8 or 16

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.

• 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.
– user99914
Oct 16, 2015 at 4:39
• 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.
– user99914
Oct 16, 2015 at 4:40
• @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? Oct 16, 2015 at 4:45
• 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 $g\in G$.
– user99914
Oct 16, 2015 at 4:55
• @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 Oct 16, 2015 at 4:59

You can try this way:

$48=2^4.3$

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)