understanding the language of a theorem The theorem is stated as follows in the book:

Let $\phi:G\rightarrow G'$ be a group homomorphism, and let
  $H=Ker(\phi)$. Let $a\in G$. Then the set 
$\phi^{-1}[\{\phi(a)\}] = \{x\in G | \phi(x)=\phi(a)\}$
is the left coset $aH$ of $H$, and is also the right coset $Ha$ of
  $H$. Consequently, the two partitions of $G$ into left cosets and into
  right cosets of $H$ are the same.

I'm trying to parse this statement and it's not clear to me what claim the author is trying to make at the very end when he says "Consequently, the two partitions of $G$ into left cosets and into right cosets of $H$ are the same." I'm under the impression that, in general, the left and right cosets are not always the same. Under what condition are they the same? Under the condition that you have a homomorphism? 
Let me mention that at this point, we're not supposed to know what a normal subgroup is. The author introduces the idea of a normal subgroup 2 pages later.

If $N \subseteq G$ is a subgroup such that $gNg^{-1} = N$ for all $g \in G$, then $N$ is called a normal subgroup. See Wiki on that. This is equivalent to saying that any left and right cosets $gN = Ng$ for any $g \in G$ are the same.
So the statement actually just says that every kernel of a homomorphism is a normal subgroup.
This is easy to see since for all $h \in H = \mathrm{ker} (\phi)$, $a \in G$ you get:
$$\phi (aha^{-1}) = \phi (a) \phi(h) \phi (a^{-1}) = \phi (a) \phi (a^{-1}) = 1,$$ so $aha^{-1} \in \mathrm{ker}(\phi) = H$.
Therefore $aHa^{-1} \subseteq H$, multiplying by $a^{-1}$ from the left and $a$ from the right gives the other inclusion for its inverse. Since $a$ was arbitrary, equality follows.
Maybe you were confused by the definition of coset: I would read $aH$ as $\{ ah \in G;\; h \in H\}$, the set of all elements of the form $ah$ with $h \in H$. Similiary for $aHa^{-1}$. That both definitions given are equivalent, is done by checking:
$$x \in aH \Leftrightarrow a^{-1}x \in H \Leftrightarrow \phi(a^{-1}x) = 1 \Leftrightarrow \phi(a) = \phi(x).$$ 
