Show that no group of order 48 is simple

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?

• Welcome to MSE! Please restate the title in the body of the post, and please define all the notation you use – 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.

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