So I get the answer as follows (which is correct I believe):

{$0$} x {$0$} $\vartriangleleft$ $\mathbb Z_7$ x {$0$} $\vartriangleleft$ $\mathbb Z_7$ x <$4$> $\vartriangleleft$ $\mathbb Z_7$ x <$2$> $\vartriangleleft$ $\mathbb Z_7$ x $\mathbb Z_{12}$

Now in order to be a composition series, the normal series:

1 = $G_0 \vartriangleleft G_1 \vartriangleleft ... \vartriangleleft G_k = G$

$G_{i+1}$ / $G_i$ must be simple $\forall i$ , where simple means its only normal subgroups are itself and the trivial group.

Maybe I am not totally understanding simple but why couldn't you start the series as;

{$0$} x {$0$} $\vartriangleleft$ $\mathbb Z_7$ x {$0$} $\vartriangleleft$ $\mathbb Z_7$ x <$6$> $... $

Thanks in advance.


For a finite group, length of composition series must be same. But composition series are not unique. You can have two different composition series for a same group.

$\{0\} \times \{0\} ⊲ \mathbb Z_7 \times \{0\} ⊲ \mathbb Z_7 \times \left<6 \right> ⊲ \mathbb Z_7 \times \left<2\right> ⊲ \mathbb Z_7 \times \mathbb Z_{12}.$

$\{0\} \times \{0\} ⊲ \mathbb Z_7 \times \{0\} ⊲ \mathbb Z_7 \times \left<6 \right> ⊲ \mathbb Z_7 \times \left<3\right> ⊲ \mathbb Z_7 \times \mathbb Z_{12}.$

The above two are also composition series of $\mathbb Z_7 \times \mathbb Z_{12}.$

Theorem: Let $G$ be non-trivial finite group. Then $G$ has a composition series. The composition factors in a composition series are unique, namely, if $1= N_0 ⊲ N_1 ⊲ N_2 ⊲ \cdots ⊲ N_r = G $ and $1= M_0 ⊲ M_1 ⊲ M_2 ⊲ \cdots ⊲ M_s = G $ are two composition series of $G,$ then $r = s$ and there is a permutation $\pi,$ of the set $\{1, 2, \cdots r \},$ such that $M_{\pi(i) + 1}/M_i \cong N_{i + 1}/N_i, 1 \leq i \leq r-1.$

  • $\begingroup$ You are welcome. :) $\endgroup$ – Krish Jan 9 '15 at 15:27
  • 1
    $\begingroup$ Just to clarify, this is the Jordan-Hölder Theorem? (The one stated). $\endgroup$ – mrhappysmile Jan 9 '15 at 17:23
  • 1
    $\begingroup$ Yes. This is Jordan-Hölder Theorem. $\endgroup$ – Krish Jan 9 '15 at 17:33

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