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?

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    $\begingroup$ Welcome to MSE! Please restate the title in the body of the post, and please define all the notation you use $\endgroup$ – leibnewtz Oct 27 '18 at 6:05

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 $S_3$.


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.

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    $\begingroup$ The intersection of all the 2-Slow subgroups need not be just $\{e\}$. It could, a priori, have order $2$ or $4$ or $8$. $\endgroup$ – Andreas Blass Oct 27 '18 at 20:47

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