Well, in the proof of the following 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,
They say that since $H\cap(C+K)= K$, we have $H\cap C=\{e\}$. But how do they have that
this part $H\cap(C+K)= K$?
Can someone prove that please is because I don't see why is that true with the arguments presented in that proof.
I have the following attempt :
We know that $G/K=(C+K)/K \cong H' \Rightarrow H \cap (C+K)=K$
If there is $x \in H \cap C, x \neq e,$ then $H \cap (C + K) = K \Rightarrow x \in K.$ But by your assumption $C\cap K = \{e\}$
more precisely :
I want to prove (convincing myself), why is this rght. In the proof of the 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
they have that since $H\cap(C+K)= K$, we have $H\cap C=\{e\}$. But how do they have the part of $H\cap(C+K)= K$? is because they say that $H'$ is the preimage of a certain map, then they can do the following:
$G/K= (C+K)/K \oplus H/K$
but I do not know if that is enough to justificate that step, or how can they have the conclusion that $H\cap(C+K)= K$,and well the part that $H\cap C=\{e\}$ is obvious from here, and in the proof of the same lemma they say that because $K$ has prime order then $K\cap C=\{e\}$, is that the only reason? Can someone prove those things please is because I don't see why is that true with the arguments presented in that proof. Thanks in advance.
I am back in the discussion and I have edited my post to be clearler :)
If there is $x \in H \cap C, x \neq e,$ then $H \cap (C + K) = K \Rightarrow x \in K.$ But by your assumption $C\cap K = \{e\}$
Let $h = c + k \in H \cap (C + K), c \neq e.$ Then $c = h - k \in H \Rightarrow c + K = h + K$ is a non-trivial element in $((C + K)/K) \cap (H/K)$ which contradicts the fact that $G/K = (C + K)/K \oplus H/K.$