Re: what's the composition series for the additive group Z/12Z?

In article <1132237895.654683.65050@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
christina505@xxxxxxxx writes:
>
>Tonico wrote:
>> christina505@xxxxxxxx wrote:
>> > Yeah, I am not sure with my answer
>> > That is, 0<(=) <2><(=)<1>=Z/12Z
>> > the generator of <2> is 2?
>> > Dr Tim wrote:
>> **********************************************888
>> Hi:
>> A composition series is a series all of which factors are simple
>> groups. Since we have here an abelian finite group, this forces the
>> factors to be of prime order. We have here:
>>
>> Z/12Z =: Z_12 := {0,1,2,3,4,5,6,7,8,9,10,11} , with the operation sum
>> modulo 12.
>>
>> <2> = {0,2,4,6,8,10}, which is NOT of prime order, and thus the factor
>> <2>/<0> ~ <2> is NOT simple., so that <0> <= <2> <= Z_12 is NOT a
>> comp. series.
>>
>> So now you have a lead: the first element "above" <0> must be a group
>> of prime order, which by Lagrange must ALSO divide the order of Z_12.
>> Try 2,3 , say. For example, 4
>> above generates a sbgp. of order 3 of Z_12...
>>
>> Good luck!
>> Tonio
>
>Thank you for your hint, Tonio!
>>>From the lattice of the sugroups of Z/12Z, I think <0><(=)<4><(=)Z/12Z
>and
><0><(=)<6><(=)Z/12Z should be composition series of Z/12Z, right?

No, neither of those are composition series, because not all of the factors
have prime order.

>What is the other composition series of Z/12Z?
>I don't think <0><(=)<3><(=)Z/12 is right because <3> is a sugroup of
>order 4 which is not a prime number.

Exactly!

Derek Holt.
>


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