I was reading Foote & Dummit's book on abstract algebra and this question arose. The authors start their discussion on quotient groups by first defining the quotient group $G/K$ where $K$ is the kernel of some homomorphism $\varphi$ defined on $G$. Let me introduce some of their definitions and terminology.
Let $G$ and $H$ be groups and $\varphi: G \to H$ a group homomorphism with kernel $K$. A fiber in $G$ over $h$ is the pre-image the singleton $\{h\} \subset H$, i.e. $\varphi^{-1}(\{h\})$. The quotient group $G/K$ is defined to be the set of all such fibers with group operation defined as follows. If $X = \varphi^{-1}(\{h\})$ and $Y = \varphi^{-1}(\{s\})$, then $XY := \varphi^{-1}(\{hs\})$.
Now, I might be wrong but as far as I know, the homomorphism $\varphi$ is not assumed to be surjective. Thus, in principle the empty set can be an element of $G/K$.
However, the authors discuss a more natural way to define $G/K$ in which the explicit use of the homomorphism $\varphi$ is not required. First, they prove that if $X$ is any fiber and $u \in X$ then $X = uK :=\{uk: \hspace{0.1cm} k \in K\}$. Then, they proceed to prove that $G/K$ is precisely the set of all such left cosets $gK$, $g \in K$. At this point, I prefer not to follow the argumentation of the book. My thoughts are the following. If $X$ is a nonempty fiber on $G/K$, then $X = uK$ for some $u \in X \subseteq G$, so that every nonempty fiber is a left coset of the form $gK$, $g \in G$. Conversely, if $g \in G$, there exists some fiber $X$ such that $g \in X$ and $X = gK$. However, the reasoning here is not taking the empty set into account and, in addition, the empty set cannot be expressed as a left coset since $K$ is a subgroup of $G$.
Question: Does the quotient group $G/K$ contain the empty set or not?
Note: I know that the problem can be solved by taking $H = \text{Im}\varphi$ instead, which is a subgroup of $H$, but there is no mention of this fact on Foote & Dummit's book and I got tricked by it.
Best Answer
No, the empty set is not an element of the quotient group $G / K$. This appears to be a mistake in Dummit and Foote: they should define $G / K$ to be set of nonempty fibers of $\varphi : G \to H$.
One explanation for defining quotient groups in this way is that we want the first isomorphism theorem to hold: if $\varphi : G \to H$ is a group homomorphism then we want $G / \ker \varphi \cong \operatorname{im} \varphi$, however if $\varphi$ is not surjective and hence with the incorrect definition $G / \ker \varphi$ contains the empty set (and $G$ is finite), then these two sets will not even have the same size.