TITLE: When k[G/H] is multiplicity free G module ? 
QUESTION [9 upvotes]: Consider finite group G and its subgroup H, and representation  of G in k[G/H] i.e. functions on G/H.
Question: What is known about the question: when k[G/H] is multiplicity free ? (Let us consider k - complex numbers for simplicity).
More generally consider $Ind^G_H V$ some induced representation of G from $H$, same question. 
(In case V= trivial representation we get previous question).
Are there some general conditions ? What is know about the cases G = S_n, A_n, GL_n(F_q) ? 
Example: $S_{n-1} \in S_n$ is  multiplicity free since G/H = C^n - standard representation.
Particular question $H=GL_{n}(F_q) \subset G= GL_{n+1}(F_q)$ is C[G/H] multiplicity free ?
(This is true from GL(C) and leads to Gelfand-Zeitlin basis theory, well , may be it is not correct). 
More generally what about induced representations in this case ? 
Morally multiplicity free means that $H$ is "big" subgroup of G, however I do not know any precise way to say what "big" means.
PS
I googled "Multiplicity-free permutation representations of the alternating groups", which promises survey on the topic, but but I have no access to this file:(
https://www.ideals.illinois.edu/handle/2142/22828
If someone can help ... would be great...

REPLY [6 votes]: The fact (mentioned in Florian Eisele's answer) that the endomorphism algebra must be commutative gives a further necessary condition for $\mathbb{C}_H\uparrow^G$ to be multiplicity free: $N_G(H)/H$ must be abelian. For $\mathbb{C} N_G(H)/H$ appears as a subalgebra of the endomorphism algebra, coming from those double cosets labelled by an element of the normalizer.
In the case when $G$ is the symmetric group $S_n$, the problem of which permutation characters are multiplicity-free is completely solved for $n \geq 66$ by work of Mark Wildon: http://arxiv.org/abs/0903.2864