TITLE: 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.
QUESTION [7 upvotes]: 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

REPLY [2 votes]: 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.