I know that finite abelian groups are solvable, because it means $ G^{(1)}=1.$ I also know that a group is solvable if you have a tower of $1=G_{s} \ \triangleleft \ G_{s-1} \ \triangleleft \ \dots \ \triangleleft \ G_1 \triangleleft \ G_0 = G$ with $G_i/G_{i+1}$ cyclic.

My intuition for the tower would be to write $1=G^{(1)} \ \triangleleft \ G$ but I don't see how $G/G^{(1)}$ is cyclic.


  • $\begingroup$ The condition about the series of groups with cyclic subgroup applies only to finite solvable groups. An infinite solvable group does not necessarily have such a series. $\endgroup$ – Derek Holt Apr 15 '17 at 21:49

$G/1$ is not necessarily cyclic.

However, the classification theorem for finite abelian groups tells us that we can write $$ G = C_1 \times C_2 \times C_3 \times \dots \times C_n,$$ where $C_1, C_2, C_3 , \dots , C_n$ are all cyclic.

So we can take $$ G_n = 1, \ \ \ \ \ G_{n-1} = C_1, \ \ \ \ \ G_{n-2} = C_1 \times C_2, \ \ \ \ \ G_{n-3} = C_1 \times C_2 \times C_3 , \ \ \ \ \ \dots \ \ \ \ \ G_0 = G.$$

Then, for each $i$, we have $$ G_{i}/G_{i+1} = C_{n-i},$$ which is cyclic.


Unless I didn't understand your question correctly, I think you got your definitions mixed up.

A solvable group is a group which has a subnormal "tower" with abelian quotients, not necessarily cyclic.

In that case, it is clear that for a finite abelian group $G$ the tower you suggested is good, as the quotient is $G$ which is abelian.

  • $\begingroup$ To be fair, I've seen both definitions being used (cyclic quotients and abelian quotients) - and of course they both work. But +1 for your answer nonetheless. :) $\endgroup$ – Kenny Wong Apr 15 '17 at 21:08
  • $\begingroup$ I was actually not aware that the cyclic definition exists and equivalent. Thanks! $\endgroup$ – The way of life Apr 16 '17 at 7:10

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