Show that no group of order 48 is simple

Question

Show that no group of order 48 is simple

I was wondering if I was allowed to do something along this line of thinking:

Let $n_2$ be the number of $2$-Sylow groups.

$n_2$ is limited to $1$ and $3$ since these are the only divisors of 48 that are equivalent to $1 \mod 2$.

$n_2=3$ (since if $n_2=1$ the group is definitely not simple)

Each $n_2$ subgroup contains 1 distinct element and there are 3 of these subgroups hence there are 3 distinct elements.

We have $48-3=45$ elements to account for.

At this point, can I assume that these 45 elements form a subgroup and then solve this proof by proving a group of 45 elements form a p-Sylow normal subgroup?

Answer 1

A Sylow $2$-subgroup contains fifteen non-identity elements. Two distinct Sylow $2$-subgroups could conceivably meet in an order $8$ subgroup, so it is difficult to count how many elements are contained in the union of the Sylow $2$-subgroups.

But there are either $1$ or $3$ Sylow $2$-subgroups. In the latter case $G$ acts on them transitively by conjugation, so there is a non-trivial homomorphism from $G$ to $S_3$.

Answer 2

I obviously did not understand what a $2$-Sylow subgroup encompasses. I've attached my corrected answer, comments are welcome. It's a bit different from the answer above but it's the same gist I think.

$48=3 \times 2^4$

$n_2=1,3$ since $1, 3$ are the only divisors of $48$ which are equivalent to $1\mod 2$.

Consider the $2$-Sylow subgroup.

$n_2\neq 1$ or else the $n_2$ subgroup will be normal to itself so $n_2=3$.

Each $2$-Sylow subgroup has $2^4$ elements and the intersection of these $2$-Sylow subgroups (assuming they're unique) is $\{e\}$ the identity.

So there are $15$ distinct element in each $2$-Sylow subgroup.

$3 \times 15=45$.

So there are $3$ distinct elements not accounted for but one of them is $e$, the identity, so there are actually only $2$ distinct elements not accounted for.

Consider the $3$-Sylow subgroup.

$n_3$ must be $1$ since there are only $3$ elements remaining, one of which is $e$.

So the $3$-Sylow subgroup conjugates to itself so it must be normal.

Answer 3

Let $G$ be the group of order $48$.

$|G| = 48 = 2^4 \cdot 3$

Then number of Sylow $2$-subgroups are $1$ or $3$. If it is one then we are done.

So assume there are $3$ Sylow $2$-subgroups. Let $H_1$ and $H_2$ be two distinct Sylow 2-subgroups and $|H_1| = 16 = |H_2|$

Now $|H_1 \cap H_2| \neq 1 \;\text{and} \;16$, [as they are distinct.]

$|H_1 \cap H_2| \neq 2 \;\text{and}\; 4$, [otherwise number of element in $H_1H_2$ will be more than 48]

So $|H_1 \cap H_2| = 8 $

I will denote $H_1 \cap H_2$ as $H$ and normalizer of $H_1 \cap H_2$ in G as $N(H)$.

My claim is that $N(H) = \{g \in G : gHg^{-1} = H\} = G$.

$H$ is normal in both $H_1$ and $H_2$.

Since $N(H)$ is the largest subgroup of $G$ to whom $H$ is normal, so $H_1$,$H_2$ are subgroup of $N(H)$.

$\implies|N(H)| \ge 16$

Also $|N(H)|> 16$, [as $H_1$ and $H_2$ are distinct]

And $16 \;\text{divides}\; |N(H)|$ & $|N(H)| \;\text{divides}\; 48,\;\;$ [as $H_1 \le N(H) \le G$]

$\implies |N(H)| = 48$

So, $G=N(H)$

As any subgroup is normal in its normalizer, so $H$ is normal in $G$.

Hence we got an proper non trivial normal subgroup of $G$.