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By Jordan-Holder thm, it is known that every finite group has a unique composition series.(Here, unique means that there is only one kinds of such series.)

And it is known also that composition series of a finite group does not determine its group. For example, $\mathbb{Z}_2 \times \mathbb{Z}_2$ and $\mathbb{Z}_4$ have the same composition series.

But $\mathbb{Z}_2 \times \mathbb{Z}_2$ has three different composition series in it and $\mathbb{Z}_4$ has only one such series.

My question arises here.

If two finite groups have the same composition series and their total number of such series are also same, then does the two groups equal?

In other words, composition series and its number in it completely determines the original group?

Andrew
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    I think $\mathbb{Z}_2$ suppose to be $\mathbb{Z}_4$. And what do you mean by "the same" composition series? – Ofir Schnabel Jul 31 '15 at 11:24
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    @Ofir, Thanks for your correction. I mean that the composition series of $G$ and $H$ are same if their length are same and each quotient facters in the series are equal acorrding to their corresponding order. $G_i/G_{i-1}\simeq H_i/H_{i-1}$. – Andrew Aug 01 '15 at 03:37

2 Answers2

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The groups $S_6$ and ${\rm PGL}(2,9)$ both have a unique composition series with normal subgroup $A_6$ and quotient group of order $2$.

Derek Holt
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  • Thanks, But how do we know $S_6$ and $PGL(2,9)$ are not isomorphic? – Andrew Aug 01 '15 at 03:39
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    Because it's well-known! It is not hard to see that ${\rm PGL}(2,9)$ has an element of order $8$ as the image of a diagonal matrix with entries $[w,1]$ for $w$ a primitive field element, but $S_6$ has no element of order $8$. There is a third non-isomorphic group with the same unique composition series known as $M_{10}$. – Derek Holt Aug 01 '15 at 08:10
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    @Holt, Thank you again for your illuminations. I leraned a lot from you. – Andrew Aug 03 '15 at 09:05
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You should except that this is definitely not true. Generally there are a ton of groups with a given set of composition factors. The number of distinct composition series does not seem like an invariant which can carry too much information about the structure of the group, so you would expect some coincidences.

One reasonable place to look for counterexamples is among solvable groups, since two solvable groups of equal order have the same composition factors.

Mikko Korhonen
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  • But, if I have calculated correctly, then $A_4$ and $D_{12}$ do not have the same number of composition series. I think $A_4$ has $3$ and $D_{12}$ has $4$. In any case, I don't think there is much point in trying to answer this question until the OP clarifies what he means by two groups having "the same composition series". – Derek Holt Jul 31 '15 at 13:02
  • @DerekHolt: Thanks for catching my mistake, you are right. But I do think there is only one reasonable interpretation to this question. – Mikko Korhonen Jul 31 '15 at 13:39
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    Unless I made a mistake again, $A_4$ and the cyclic group of order $12$ both have three distinct composition series. – Mikko Korhonen Jul 31 '15 at 13:45
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    I can think of two interpretations: the same composition factors, or the same composition factors in the same order. For the second one, both groups would have to have the same number of compostion series for each possible order of composition factors within the series. But it's not hard to find examples! – Derek Holt Jul 31 '15 at 13:52
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    Yes that's right, but the orders of the composiution factors within the compostiion series do not correspond exactly, because in all three series for $A_4$, the fact or order $3$ comes at the top of the group. – Derek Holt Jul 31 '15 at 16:36