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I have a question which is: find the composition factors of a group of order $48$. But I can't see how I can do this.

If I choose my group to be $C_{48}$ then it has composition series : $$\{1\}\triangleleft C_2 \triangleleft C_4 \triangleleft C_8 \triangleleft C_{16} \triangleleft C_{48}$$ So the composition factors are $C_3$ and four $C_2$.

But then if I choose a different group from Sylow theory I have that there is either a normal subgroup of order $16$ or $8$. So if it is not $16$, it is $8$ so call it $N$, then we have a subnormal series with $\{1\} \triangleleft N \triangleleft G$ where any refinment will be to the left of $N$ so we have a composition factor of size 6?

Have I done something wrong?

Thanks for any help

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Any group of order 6 has a normal subgroup of order 3, so no group can have a composition factor of order 6. Composition factors are always simple groups. The abelian simple groups are all cyclic of prime order. The smallest nonabelian simple group is the alternating group $A_5$ of order 60, so all composition factors of a group of order 48 must be cyclic of prime order. – Derek Holt Dec 22 '12 at 16:17
up vote 2 down vote accepted

The composition factors are simple groups (by definition). Now, since the group has order $48$ it is solvable (all groups of order $< 60$ are solvable), and this implies that all composition factors of it must also be solvable.

Any simple solvable group will have prime order (and any group of prime order is simple and solvable). This means that if $G$ is a solvable group and $p_1,p_2,\dots,p_k$ are the primes dividing the order of $G$, then the composition factors of $G$ are precisely the cyclic groups of orders $p_1,p_2,\dots,p_k$.

In you specific case you then get that the composition factors are precisely the cyclic group of order $2$ and the cyclic group of order $3$.

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