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**frankdent1** **Prove there are no simple groups of order 48**

Ok, here's what i have so far.

$\displaystyle 48=2^4*3$

$\displaystyle n_{2} = 1$ mod $\displaystyle 2$ & $\displaystyle n_{2} | 3$

$\displaystyle n_{2} = 1$ or $\displaystyle 3$

$\displaystyle n_{3} = 1$ mod $\displaystyle 3$ & $\displaystyle n_{3} | 16$

$\displaystyle n_{3} = 1, 4,$ or $\displaystyle 16$

So we must eliminate these cases to get $\displaystyle n_{2} = 1$ or $\displaystyle n_{3} = 1$:

$\displaystyle (n_{2}, n_{3})=(3,4)$

$\displaystyle (n_{2}, n_{3})=(3,16)$

How do I eliminate these cases?