The usual [second isomorphism theorem](https://en.wikipedia.org/wiki/Isomorphism_theorem#Second_isomorphism_theorem) for groups is: let $G$ be a group, $S$ and $N$ subgroups with $N$ normal, then $SN$ is a subgroup of $G$, $S\cap N$ is a normal subgroup of $S$ and $SN/N \simeq S /S \cap N$.  

($N$ need not to be a normal subgroup as long as $S$ is a subgroup of the normalizer of $N$).


Now if $S$ **and** $N$ are normal subgroups of $G$ then $SN/N \simeq S /S \cap N$ **and** $SN/S \simeq N /S \cap N$, so that the normal chains $(S \cap N \subset S \subset SN)$ and $(S \cap N \subset N \subset SN)$ are equivalent.


It's this last property that we would like to generalize to the inclusions of groups $(H \subset G)$.


The notion of *normal* subgroup $K \triangleleft G$ (i.e. $ \forall g \in G \text{ , } Kg=gK $) can be generalized by the notion of *normal intermediate* subgroup (motivated by the prop.3.3 p476 of [this paper](https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-50/issue-2/Normal-intermediate-subfactors/10.2969/jmsj/05020469.full)).


**Definition**: $K$ is a **normal intermediate** subgroup of the inclusion $(H \subset G)$ if $H \subset K \subset G$, and $$\forall g \in G \text{ , } KgH=HgK $$
**Remark**: If $K\_i$ is a normal intermediate subgroup of $(H \subset G)$, then so are $K\_1 \cap K\_2$ and $K\_1K\_2$,  

and if $H \subset H\_0 \subset K\_i \subset G\_0 \subset G$ then $K\_i$ is also a normal intermediate subgroup of $(H\_0 \subset G\_0)$.


**Definition**: A chain of subgroups $(K\_1 \subset K\_2 \subset \dots \subset K\_n)$ is a *normal intermediate* chain if $K\_i$ is a normal intermediate subgroup of the inclusion $(K\_{i-1} \subset K\_{i+1})$.


**Definition**: Two chains of subgroups $(K\_1 \subset K\_2 \subset \dots \subset K\_n)$ and $(L\_1 \subset L\_2 \subset \dots \subset L\_m)$ are equivalent if $m=n$ and $\exists \sigma \in S\_{n-1}$ such that $(K\_i \subset K\_{i+1}) \sim (L\_{\sigma(i)} \subset L\_{\sigma(i)+1})$.


**Definition**: Two inclusions of groups are equivalent, $(A \subset B) \sim (C \subset D)$, if: $(A/A\_B \subset B/A\_B) \simeq (C/C\_D \subset D/C\_D)$ with $A\_B$ the [normal core](https://groupprops.subwiki.org/wiki/Normal_core) of $A$ in $B$.



> 
> **Question**: Let $K$ and $L$ be normal intermediate subgroups of an inclusion $(H \subset G)$ then, are the *normal intermediate* chains
> $(K \cap L \subset K \subset KL)$ and $(K \cap L \subset L \subset KL)$ equivalent ?
> 
> 
> 


**Examples**: See [this post](https://mathoverflow.net/questions/158662/normal-intermediate-subgroup-and-normal-core) and note that [this answer](https://mathoverflow.net/questions/158662/normal-intermediate-subgroup-and-normal-core/159108#159108) gives here an example.


**Motivation**: A Jordan-Hölder theorem generalized to the inclusions of groups.



A second isomorphism theorem for the inclusions of groups
**No**, the following theorem gives counterexamples: 



> 
> **Theorem:** $\forall n \ge 3$, $A\_{n+1}$ and $S\_n$ are *normal intermediate subgroups* of the inclusion $(A\_n \subset S\_{n+1})$, but $(A\_n \subset S\_n \subset S\_{n+1} )$ and $(A\_n \subset A\_{n+1} \subset S\_{n+1} )$ are not equivalent.
> 
> 
> 


**Lemma**: $\forall n \ge 3$, $S\_{n}$ is a *normal intermediate subgroup* of the inclusion $(A\_n \subset S\_{n+1})$. 


*Proof*: The equality $A\_n . g . S\_n = S\_n . g . A\_n$ is clear if $g \in S\_n$, but $S\_{n+1} = \langle S\_n , \tau \rangle$ with $\tau=(n,n+1)$, so it suffices to prove that $A\_n . \tau . S\_n = S\_n . \tau . A\_n$, **because** $A\_{n} \triangleleft S\_{n}$:  

If $\sigma$, $\sigma' \in S\_n$ then $a\_1\sigma\tau\sigma's\_1=\sigma a\_2 \tau s\_2 \sigma=\sigma s\_3 \tau a\_3 \sigma = s\_4\sigma\tau\sigma' a\_4$ ($a\_i \in A\_n$, $s\_i \in S\_n$). 


Now $S\_n= \langle (1,2, \dots , n) , (1,2)\rangle$, and $A\_n . (n,n+1) . (1,2) = (1,2) . (n,n+1) . A\_n$  

for all $n \ge 3$, so it suffices to show that $A\_n . (n,n+1) . (1, \dots , n) \subset S\_n . (n,n+1) . A\_n$. 


But $A\_n . (n,n+1) . (1, \dots , n) = (n,n+1) . (1, \dots , n) . A\_n$,  

and $(n,n+1) . (1, \dots , n) = e.(n,n+1) . \sigma\_1 = (1,2) . (n,n+1) . \sigma\_2$,  

with $e, (1,2) \in S\_n$ and $\sigma\_1 = (1, \dots , n) \in A\_n$ if $n$ odd, $\sigma\_2 = (1,2). \sigma\_1 \in A\_n$ if $n$ even. 


(Idem, $(1, \dots , n) . (n,n+1) . A\_n \subset A\_n . (n,n+1) . S\_n $)  

So, $A\_n . (n,n+1) . S\_n = S\_n . (n,n+1) . A\_n $ and then $A\_n . g . S\_n = S\_n . g . A\_n$ $\forall g \in S\_{n+1}$. $\square$ 


*Proof of the theorem*: $A\_{n+1}$ is obviously a normal intermediate subgroup because $A\_{n+1} \triangleleft S\_{n+1}$.  

$S\_n$ is also a normal intermediate by the previous lemma. Next $(A\_n \subset A\_{n+1}) \not\sim (S\_n \subset S\_{n+1})$. $\square$ 


**Remark**: For correctness (with the question) note that $A\_{n+1} \cap S\_n = A\_n$ and $A\_{n+1}.S\_n = S\_{n+1}$. 


