Clarifying the First Isomorphism Theorem In the first Isomorphism theorem it says that G/ker(f) is isomorphic to Im(f). As G is the set of left cosets of ker(f) in G, and as the ker(f) contains the identity element e, doesn't this mean that G/ker(f) = G? In which case wouldn't it only be isomorphic to Im(f) if f was a monomorphism? 
I know there's something wrong with my reasoning but I can't figure out what! Please explain it to me like I'm 5, finding this really hard to get my head around.
 A: Be careful. You're saying that "$G$ is the set of left cosets of $ker(f)$ in $G$". Actually, $G$ is the disjoint union of the left cosets of $ker(f)$ in $G$, since these cosets are the equivalence classes of the congruence modulo $ker(f)$ relation on $G$. It is $G/ker(f)$ that is the set whose elements are the left cosets of $ker(f)$ in $G$. So, the equality $G/ker(f) = G$ doesn´t make sense with the standard definitions at all. 
There is one case where one could write such an expression as an abuse of notation: if $ker(f) = \{e\}$ then the congruence modulo $ker(f)$ relation is the trivial one, i.e the left cosets of $ker(f)$ in $G$ (or the equivalence classes of the congruent modulo $ker(f)$ relation on $G$) are of the form $\{g\}$ for all $g \in G$. To see this, just pick an arbitrary element $g \in G$ and calculate its left coset. But this is $g\{e\}$ in our case, which is just $\{g\}$! So in this case $G/ker(f) =G/\{e\}= \{\{g\}, g \in G\}$ which can be seen as a structural copy of the group $G$. In fact they are isomorphic groups (where $G/ker(f)$ has the usual well defined product of left cosets, since ker(f) is always a normal subgroup of $G$). So one could write $G/ker(f) = G$ in this case, by abuse of notation. 
