Normal series and composition series

[fishturtle1]

Homework Statement

Let $G$ and $H$ be finite groups. If there are normal series of $G$ and $H$ having the same set of factor groups, then $G$ and $H$ have the same composition factors.

Relevant Equations

A normal series of a group $G$ is
$$G = G_0 \ge G_1 \ge G_2 \ge \dots \ge G_n = 1$$
where $G_i \trianglelefteq G_{i+1}$ for all $i$. The factor groups of this series are $G_{i}/G _{i+1}$ for all $i$. The length of this series is the number of distinct factors.

A composition series is a normal series whose factors are simple or trivial.

Attempt so far: We're given that $G$ and $H$ have equivalent normal series
$$G = G_0 \ge G_1 \ge \dots \ge G_n = 1$$
and
$$H = H_0 \ge H_1 \ge \dots \ge H_n = 1$$

We can assume they have the same length because they are equivalent. I think from here I need to construct two composition series; one for $G$ and one for $H$, and then show they are equivalent. By Jordan Holder theorem, any two composition series of a group will have the same factors. So this would be enough, I think.

But I'm having trouble constructing a composition series, starting with $G$... I think I need to write
$G = G_0 \ge G_{0, 1} \ge G_{0, 2} \ge \dots \ge G_{0, k} \ge G_1 \ge G_{1, 1} \ge \dots$, but I'm not sure how to define $G_{i, j}$ to such that the series is a composition series. Can I have a hint, please?

Edit: For each $G_i/G_{i+1}$ there is an $H_j/H_{j+1}$ such that $G_i/G_{i+1} \cong H_j/H_{j+1}$.
By correspondence theorem, there is a bijection between the set of subgroups of $G_i$ that contain $G_{i+1}$ and the set of subgroups of $G_{i}/G_{i+1}$. Similarly, there is a bijection between the set of subgroups of $H_i$ that contain $H_{i+1}$ and the set of subgroups of $H_i/H_{i+1}$. Hence, there is a bijection between the set of subgroups of $G_i$ that contain $G_{i+1}$ and the set of subgroups of $H_{i}$ that contain $H_{i+1}$.

 

[fresh_42]

You should note why composition series exist in your case.

We have isomorphic normal series. We need to show that their refinements lead to isomorphic composition series: The factors are in both, so you can add all refinements of either at the according place of the other. You only need to map the factors accordingly. Or use the given bijection of the factors, refine them one by one, and copy the refinement from $G$ into the series of $H$.

Do you have an example for $G$ and $H$ which fulfill the conditions of the exercise, and which are not isomorphic?

 

[fishturtle1]

Thanks for the reply!

Proof: We are given that $G$ and $H$ are finite which implies they both have composition series. We are also given that there exists normal series
$$G = G_0 \ge G_1 \ge \dots \ge G_n = 1$$
and
$$H = H_0 \ge H_1 \ge \dots \ge H_n = 1$$
that are equivalent. Hence, for each $G_i/G_{i+1}$ there exists $H_{j}/H_{j+1}$ such that $G_i/G_{i+1} \cong H_j/H_{j+1}$, and vice versa. Fix $i$. By Correspondence theorem, there is a bijection $f$ between the set of subgroups of $G_i$ that contain $G_{i+1}$ and the set of subgroups of $H_j$ that contain $H_{+1}$.

Consider the normal series $G_i \ge G_{i+1}$. We will refine it into a composition series. If it already a composition series, we are done. Otherwise, choose maximal normal subgroup $G_{i, 1} \le G_i$ and consider the normal series $G_i \ge G_{i, 1} \ge G_{i+1}$. By choice, $G_i/G_{i, 1}$ is simple. If $G_{i, 1}/G_{i+1}$ is simple, we are done. Otherwise, choose maximal normal subgroup $G_{i, 2} \le G_{i, 1}$. If
$G_i \ge G_{i, 1} \ge G_{i, 2} \ge G_{i+1}$ is a composition series, then we are done. Otherwise, choose maximal normal subgroup $G_{i, 3} \le G_{i, 2}$ and consider $G_i \ge G_{i, 1} \ge G_{i, 2} \ge G_{i, 3} \ge G_{i +1}$. ... Continuing in this way, we get
$$G_i \ge G_{i, 1} \ge G_{i, 2} \ge \dots \ge G_{i, k} \ge G_{i+1}$$
is a series whose factors are all simple.

It follows ? that
$$H_j \ge f(G_{i, 1}) \ge f(G_{i, 2}) \ge \dots \ge f(G_{i, k}) \ge H_{j+1}$$ is a series whose factors are simple. Moreover, these two series are equivalent? Doing this for all $i$, we get a composition series for $G$ and a composition series for $H$ that are equivalent.

We can conclude the composition factors of $G$ and the compositions factors of $H$ are the same. []

Is this on the right track?

Also: $S_3 \not\cong C_6$ but $S_3 \ge C_3 \ge 1$ and $C_6 \ge C_3 \ge 1$ are equivalent normal series.

 

[fresh_42]

A bit too long, me thinks. I just looked up how van der Waerden had explained it. It took him 4 lines: idea and formal proof are almost the same. Also, don't we have the same factors, i.e. $G_i/G_{i+1} = H_j/H_{j+1}$, so any mapping only takes place on the indices $1,\ldots,n$ and not on the group level, in my example here $f(i)=j$?

 

[fishturtle1]

Yes, I think you're right about the mapping taking place on the indices.

So if $G_i/G_{i+1} = H_j/H_{j+1}$, then for any normal series $G_i \ge G_{i,1} \ge \dots \ge G_{i, k} \ge G_{i+1}$ there exists equivalent normal series $H_j \ge H_{j, 1} \dots \ge H_{j, k} \ge H_{j+1}$? I'm sorry if it's obvious, but why is this true?

edit: never mind! I think i see why... thinking about it...

 

[fishturtle1]

I tried reading van der Waerden's proof and just want to make sure I understand... Suppose we have $G_i/G_{i+1} = H_j/H_{j+1}$. There is a correspondence between the normal series for $G_i/G_{i+1}$ and the series' from $G_{i}$ to $G_{i+1}$.
Hence, for any refinement of a series from $G_i$ to $G_{i+1}$, there is a corresponding normal series for $G_i/G_{i+1}$. Then there exists equivalent normal series for $H_j/H_{j+1}$ which has corresponding series from $H_j$ to $H_{j+1}$.

Hence, for any refinement of a series from $G_{i}$ to $G_{i+1}$, there exists equivalent series from $H_j$ to $H_{j+1}$.

 

[fresh_42]

Yes. If $G_{i}/G_{i+1}$ is not simple, then it has a proper normal subgroup. This looks like $G_{i,1}/G_{i+1}$ with $G_{i,1}\leq G_{i}$. Thus we get $G_{i} \triangleright G_{i,1} \triangleright G_{i+1}.$ We also have
$$
G_{i}/G_{i,1} \cong (G_{i}/G_{i+1})/(G_{i,1}/G_{i+1}) = (H_{j}/H_{j+1})/(G_{i,1}/G_{i+1})
$$
So $G_{i,1}/G_{i+1}$ is a normal subgroup of $H_{j}/H_{j+1}$ and thus of the form $H_{j,1}/H_{j+1}.$ This means $H_j \triangleright H_{j,1} \triangleright H_{j+1}.$ And if $G_{i,1}/G_{i+1} = H_{j,1}/H_{j+1}$ then all new factors are equal and we can proceed with the next non simple factor group.

 

[fishturtle1]

Yes. If $G_{i}/G_{i+1}$ is not simple, then it has a proper normal subgroup. This looks like $G_{i,1}/G_{i+1}$ with $G_{i,1}\leq G_{i}$. Thus we get $G_{i} \triangleright G_{i,1} \triangleright G_{i+1}.$ We also have
$$
G_{i}/G_{i,1} \cong (G_{i}/G_{i+1})/(G_{i,1}/G_{i+1}) = (H_{j}/H_{j+1})/(G_{i,1}/G_{i+1})
$$
So $G_{i,1}/G_{i+1}$ is a normal subgroup of $H_{j}/H_{j+1}$ and thus of the form $H_{j,1}/H_{j+1}.$ This means $H_j \triangleright H_{j,1} \triangleright H_{j+1}.$ And if $G_{i,1}/G_{i+1} = H_{j,1}/H_{j+1}$ then all new factors are equal and we can proceed with the next non simple factor group.

That makes sense, thank you!