Why do a coset is isomorphic to a certain set.

Question

I have encountered with the proof of the next lemma

suppose G is a finite abelian p-group, and let $C$ a cyclic subgroup of maximal order, then $G=C\oplus H$ for some subgroup $H$

at http://torus.math.uiuc.edu/jms/m317/handouts/finabel.pdf, but in the proof the say that

$(C+K)/K \cong C$

but the reason they give is only the following:

Because K has prime order, K∩C={e}, which implies that $(C+K)/K \cong C$ , but Can you proof in detail that implication please, is because I don't see why that is true, thank you a lot. I'll appreciate your answers.

Is there an alternative proof of the one offer by @Hayden, thanks

Answer 1

The Second* Isomorphism Theorem (for Groups) states that for a group $G$, $S$ a subgroup of $G$, and $N$ a normal subgroup of $G$, then

1. $SN=\{sn \in G \mid s\in S, n\in N\}$ is a subgroup of $G$,
2. $S\cap N$ is a normal subgroup of $S$, and
3. $(SN)/N\cong S/(S\cap N)$.

Thus, taking $G$ to be abelian (so we will use additive notation rather than multiplicative notation), we see that $$\frac{C+K}{K}\cong \frac{C}{C\cap K}=\frac{C}{\{e\}}\cong C.$$

(*: The ordering of the isomorphism theorems are not altogether standard, so this might be different than the numbering you might have seen before. I'm using the numbering from wikipedia for simplicity.)