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A supersolvable group is one with a normal series such that each factor is cyclic. In particular the last non-trivial term in the series has to be cyclic. But if we take take two of the same groups, the product of the last two terms wouldn't be cyclic. So the direct product ought to not be supersolvable. I don't understand what I'm missing here.

Mark
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1 Answers1

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Say $A$ and $B$ are supersolvable with normal series $1=N_0\leq N_1\leq ...\leq N_i=A$ and $1=M_0\leq M_1\leq ...\leq M_j=B$. Then the series $$1=N_0\leq N_1\leq ...\leq N_i=A\leq A\times M_1\leq A\times M_2\leq ...\leq A\times M_j=A\times B$$ is a normal series of $A\times B$ with every quotient is cyclic.

Edit : What you are missing is that the series you need to consider for $A\times B$ is not formed by taking the product of the terms in the series of $A$ and $B$. In this case, as you said, the quotients will not be cyclic.

Levent
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