Let $G$ be a finite p-primary abelian group. If a is an element of largest order in G, then $A= \langle a \rangle$ is a direct summand of G. I was trying to read the proof from Advanced Modern Algebra (Rotman), but there was something that seemed confusing to me. It's only the last part that's confusing, but I put the whole proof anyway.  

I'm not sure if I understood "$A \cap B \subseteq A \cap ((A + C') \cap B) \subseteq A \cap C' = \{0\}$". Normally, I would think that for any two sets $A$ and $B$, $A \cap B \subseteq A$ and $A \cap B \subseteq B$, because intersection can only make a set smaller (or equal to) but not bigger, right? But here it seems (at least to me) to go backwards. We need to assume that $B \subseteq (A+C') \cap B$. 
I don't understand $A \cap ((A + C') \cap B) \subseteq A \cap C' = \{0\}$ either. Here, we have to assume that $(A+C') \cap B \subseteq C'$. However, we know that the intersection $(A+C') \cap B$ for sure contains $C'$ since $C' \subseteq A+C'$ and $C' \subseteq B$. So it can only be greater than or equal to $C'$, right? 
Thank you in advance 
 A: The first inclusion $A\cap B\subseteq A\cap((A+ C')\cap B)$ is because $A\cap B\subseteq A$ and $A\cap B\subseteq(A+C')\cap B$.
Then we can prove that $((A+ C')\cap B)\subseteq C'$. That would give the second inclusion.
Assume that $g$ is in $A+C'$ and also in $B$. Then if $\pi:G\to G/C'$ is the canonical projection, we have $\pi(g)\in(A+C')/C'$ and $\pi(g)\in B/C'$. But $(A+C')/C'\cap B/C'=\{0\}$, so $g$ is in the kernel of $\pi$, which is just $C'$.
A: You want to prove that $A \cap B \subseteq A \cap ((A + C') \cap B)$ so you need to show that $A \cap B \subseteq A$ and $A \cap B \subseteq (A + C') \cap B$.  The first you have already noted is true.  For the second you need to show that $A \cap B \subseteq A + C'$ and $A \cap B \subseteq B$.  The first follows from $A \cap B \subseteq A \subseteq A + C'$ and the second you already have.
Now $A \cap ((A + C') \cap B) \subseteq A \cap C'$ is a little tricky.  It follows from
$$G/C' = (A + C')/C' \oplus B/C'$$
If $x \in (A + C') \cap B$ then because the sum is direct $x$ represents the identity in $G/C'$, so $x \in C'$.  This gives $(A + C') \cap B \subseteq C'$.  Intersecting with $A$ gives $A \cap ((A + C') \cap B) \subseteq A \cap C'$ as desired.
A: 
I'm not sure if I understood "$A \cap B \subseteq A \cap ((A + C') \cap B) \subseteq A \cap C' = \{0\}$". 

Clearly $A \subseteq A + C'$. This explains $A \cap B \subseteq A \cap (A + C') \cap B$.
Now since $G/C'$ is the direct sum of $(A + C')/C'$ and $B/C'$, we have $(A + C') \cap B \subseteq C'$.
