Do quotient groups contain the empty set? I was reading Foote &amp; 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 &amp; Dummit's book and I got tricked by it.
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.
The quotient group cannot be empty because it contains the identity element. It may be the only element but it's still at least a singleton.