Edit 2: Please note the new, more specific version of the question: [Does there always exist an irreducible representation occurring with multiplicity one when inducing from $M=Z\_K(A)$ to $K$?](https://mathoverflow.net/questions/302308/does-there-always-exist-an-irreducible-representation-occurring-with-multiplicit)


Let $K$ be a compact connected Lie group and $M\subset K$ a closed (hence compact Lie) subgroup, not necessarily connected. Let $\tau$ be an irreducible (hence finite-dimensional) complex $M$-representation. Does there exist an irreducible $K$-representation $\sigma$ such that $\tau$ occurs with multiplicity one in $\sigma|\_M$?


By Frobenius reciprocity, this is equivalent to the question: 


Does the $K$-representation $\mathrm{Ind}^K\_M(\tau)$ contain an irreducible subrepresentation that occurs in it only once?




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Edit: As a motivation, I am reading a thesis at the moment which says that 
for $(K,M)=(\mathrm{SO}(n+1),\mathrm{SO}(n))$ or $(K,M)=(\mathrm{SU}(n+1),\mathrm{S}(\mathrm{U}(1)\times \mathrm{U}(n)))$, one has even the stronger statement that
$$
[\sigma|\_M:\tau]\leq 1
$$
for any pair of irreducible representations $(\sigma,\tau)$ of $K$ and $M$. In particular, for such pairs of groups, the answer to my question is positive according to the thesis. However, it does not cite references where I could directly check this, or where I could find an answer to my question.



Does there always exist an irreducible representation occurring with multiplicity one when inducing from a closed subgroup to a compact Lie group?
Let $K=SU(2)$, $M=Z(SU(2)) = \{\pm I\}$, and $\tau$ the sign representation of $M$. Then given an irreducible representation $\sigma$ of $SU(2)$, $\sigma|\_M$ is necessarily a character with multiplicity $dim(V)$. Moreover, the sign character appears exactly for the even dimensional representations (i.e. those of odd highest weight). Thus $\tau$ always appears with even multiplicity in any irreducible representation.


Regarding your motivation, those particular pairs of groups are special - I think the key word here is Gelfand pair.


