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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.

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  • $\begingroup$ $G/\ker f$ consists of equivalence classes. $e$ is in one of these classes, but I don't see how you get from there to $G/\ker f = G$. $\endgroup$ – Hoot Jan 8 '15 at 20:48
  • $\begingroup$ My reasoning is that ker(f) contains {e} as f is a homomorphism, so the set of left cosets of it would contain the set of left cosets of {e} which would be G I think? $\endgroup$ – user2973447 Jan 8 '15 at 20:51
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    $\begingroup$ @user2973447: First, try (for your own benefit) to rewrite your entire question without using the pronoun "it". (I'm perfectly serious; "it" leads to needless confusion. I'm not entirely sure what you're asking because of "it".) Second, do you have a "favorite" non-injective homomorphism $f:G \to H$, and can you calculate $G/\ker(f)$? (If you don't have a favorite, how about taking $G$ and $H$ to be the additive group of integers, and $f(x) = 0$ for all $x$.) $\endgroup$ – Andrew D. Hwang Jan 8 '15 at 21:00
  • $\begingroup$ The cosets of $\ker f$ are much larger, in general. If you have subgroups $H \subseteq K \subseteq G$ then it is true that the (left, say) cosets of $K$ in $G$ are unions of cosets of $H$ in $G$, but that doesn't seem to present a problem. Think about $2\mathbf{Z} \subset \mathbf{Z}$. This has two cosets $2\mathbf{Z} = \{\dots, -2, 0, 2, 4, \dots\}$ and $1 + 2\mathbf{Z} = \{\dots, -1, 1, 3, \dots\}$. Neither is a coset of $\{0\}$. $\endgroup$ – Hoot Jan 8 '15 at 21:04
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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.

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