Constructing a Normal Series for G from Given Normal Series for N and G/N

[Chaos2009]

Homework Statement

Suppose that [itex]N \triangleleft G[/itex]. Show that given normal series [itex]S[/itex] for [itex]N[/itex] and [itex]T[/itex] for [itex]G / N[/itex] one can construct a normal series [itex]U[/itex] for [itex]G[/itex] such that the first part of [itex]U[/itex] is isomorphic to [itex]S[/itex] and the rest is isomorphic to [itex]T[/itex].

Homework Equations

This is from the last couple of weeks of an undergraduate Abstract Algebra course. The teacher assigned it as homework while discussing a proof of the Jordan-Holder theorem.

The Attempt at a Solution

I'd like to simply construct [itex]U[/itex] from [itex]S[/itex] and [itex]T[/itex]. Using [itex]S[/itex] would be straightforward as this is already a normal series from [itex]\left\{ e \right\}[/itex] to [itex]N[/itex]. However, I'd hoped to use correspondence theorem to map the normal series [itex]T[/itex] to a normal series from [itex]N[/itex] to [itex]G[/itex]. I believe, however that there is a problem with the part where it says this part of the series should be isomorphic to [itex]T[/itex].

 

[morphism]

What does it mean to say that two normal series are isomorphic?

 

[Chaos2009]

I guess that was the part I was confused about as well. My roommate now informs me that we defined two normal series to be isomorphic as follows:

Series [itex]S[/itex] and [itex]T[/itex] are isomorphic if there exists a bijection from the factors of [itex]S[/itex] to the factors of [itex]T[/itex] such that the corresponding factors are isomorphic.

So, that makes a bit more sense to me now.

 

[morphism]

Typically that's referred to as "equivalence", but anyway, your idea does work, i.e. it will produce an equivalent normal series.