If one subgroup of index $k$ is maximal normal in a join, are all subgroups of index $k$ maximal normal in this join?

Question

I have been self-studying abstract algebra from Dummit & Foote for the past few months and recently learnt about the Jordan-Holder theorem. I was looking at this theorem from a subgroup lattice perspective and it suggests that no matter which path we choose to walk down from the top to the bottom, the length of the path will be the same as long as the normality and maximality are ensured (please correct me if this intuition is incorrect).

This lead me to wonder - if one subgroup of index $k$ is maximal normal in a join, are all subgroups of index $k$ maximal normal in this join?

I've looked at the following examples and all of them seem to satisfy this

1. In $V_4$: $\langle a\rangle, \langle b\rangle, \langle c\rangle$ and their join $V_4$
2. In $D_8$: $\langle s,r^2\rangle, \langle r\rangle, \langle rs,r^2\rangle$ and their join $D_8$; $\langle s\rangle, \langle r^2s\rangle, \langle r^2\rangle$ and their join $\langle s, r^2\rangle$ and $\langle r^2\rangle, \langle rs\rangle, \langle r^3s\rangle$ and their join $\langle rs,r^2\rangle$
3. $Q_8$: $\langle i\rangle, \langle j\rangle, \langle k\rangle$ and their join $Q_8$

Is there something more to it or is it mere coincidence? If the former, how could I go about proving this?

To reiterate my question : If one subgroup of index $k$ is maximal normal in a join, are all subgroups of index $k$ maximal normal in this join?

A few follow up questions I had

1. Relaxing the join condition, if one subgroup of index $k$ is maximal normal, are all subgroups of index $k$ maximal normal?
2. Relaxing the maximal condition, if one subgroup of index $k$ is normal in a join, are all subgroups of index $k$ normal in this join?
3. Relaxing both join and maximal condition, if one subgroup of index $k$ is normal, are all subgroups of index $k$ normal?

Any help would be appreciated.