Assume $G$ is a finite group and $H$ a subgroup. Is it true that the number of irreducible representations of $G$ is always larger than (or equal to) the number of irreducible representations of $H$?

Here is a simple example. As mentionned by Gjergji, this is a question about the number of conjugacy classes. Take $G=\frak A_4$, which has $3$ classes (the identity, the double transpositions and the $4$cycles). Now take $H$ the subgroup spanned by a $4$cycle. Because $H=4$ and $H$ is abelian, it has $4$ classes. Edit. This is incorrect: $H$ is not a subgroup of $\frak A_4$, because a $4$cycle is an odd permutation. I apologize. Note that I an accepted answer can not be deleted. 


The number of irreducible representations equals the number of conjugacy classes, and the number of conjugacy classes in a subgroup may be more than in the whole group. 


No, an easy example is when $G= {\bf Z}/2{\bf Z} \ltimes ({\bf Z}/p{\bf Z})^2$ where the action is by switching the factors and $H = ({\bf Z}/p{\bf Z})^2$. $G$ has $2p$ 1dimrepresentations and $(p^2p)/2$ 2dim representations which for big $p$ is smaller than the number of representations for $H$ which is $p^2$. 


Dihedral groups also give examples. More generally, for a subgroup $H$ of a finite group $G$, the sum of the $p$th power of dimensions of representations of $H$ is at most that for $G$, provided $p\geq 1$ (here $p$ is not a prime, but one should think of $L^p$norms, including the limit as $p$ goes to infinity, which corresponds to the maximal dimension.) This monotony property fails for any $p$ in the interval $[0,1[$ (again, dihedral groups give examples). 


In many cases, the result is rather in the opposite direction. Let $k(G)$ denote the number of conjugacy classes of a finite group $G$. Now take $G$ to be be a semdirect product of the form $VM$, where $V$ is an elementary Abelian $p$group and $M$ is a group of linear transformations of order prime to $p$ of $V$. Then $k(G) \leq V$, and it is rare for equality to be achieved ( see Gluck, David; Magaard, Kay; Riese, Udo; Schmid, Peter The solution of the $k(GV)$problem. J. Algebra 279 (2004), no. 2, 694–719). The precise conditions under which equality is achieved have not been characterized to date as far as I know), so taking $H = V$ usually leads to $k(G) < k(H)$ (and always to $k(G) \leq k(H)$) for this choice of $H$. For an explicit example, if $V$ is elementary Abelian of order $p^n$ for some odd prime $p$, and $M$ is cyclic of order $(p^n 1)/2$, acting as the square of a Singer cycle on $V$, then $G = VM$ has $2 + (p^n  1)/2$ conjugacy classes, so this is less that $p^n$ unless $p^n = 3.$ While the result of Gluck,Magaard,Riese and Schmid and some of its prerequisites rely on the classification of finite simple groups, many cases can be proved without the classification, for example, the case where $G$ is solvable. 

