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I have question that asks me to find the composition series of $C_p\times C_p$, now these are all isomomrphic to the series $\{1\}\lhd C_p \lhd C_p\times C_p$ but the questions wants all the series explicitly and so my solution was as follows:

Let $G=<g>\times <h>$ then the composition series are all of the form:

$\{1\}\lhd\ <g^ih^j> \ \lhd C_p\times C_p$ for $i,j\in \{1,\ldots p\}$ which gives me $p^2$ of these but apparently there are only $p+1$ of these? Where have I gone wrong?

john smith
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  • Some of those subgroups are the same, and one of them is the trivial subgroup. Think about when two elements generate the same subgroup. – Tobias Kildetoft Mar 03 '14 at 11:42
  • @TobiasKildetoft but don't they all generate the same subgroup $C_p$? – john smith Mar 03 '14 at 11:46
  • They generate isomorphic subgroups (apart from the one where you take $i=j=p$), but not all the same subgroup. – Tobias Kildetoft Mar 03 '14 at 11:47
  • @TobiasKildetoft oh I think I get what you mean in that $=<g^2h^2>...$ and other things in that if i wrote these out they would be exactly the same (not just isomorphic)? – john smith Mar 03 '14 at 11:47
  • @TobiasKildetoft so I want everything of the form $<g^ih>$ for $i\in {0, \ldots p}$ and $$ which gives the $p+1$ – john smith Mar 03 '14 at 11:49
  • Right. An easy way to check that the number of them should be $p+1$ is to note that there are $p^2-1$ elements of order $p$, and each subgroup of order $p$ contains $p-1$ of them. – Tobias Kildetoft Mar 03 '14 at 13:28

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