Why does $\langle \bar a \rangle \cap \overline K = \{\bar e \}$?

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Theorem $:$ Let $G$ be a finite abelian $p$-group. Let $a$ be an element of maximal order in $G.$ Then $\exists$ a subgroup $K$ of $G$ such that $$G \cong \langle a \rangle \times K.$$

I have read the proof of this theorem from this Proof Wiki article. At the very last paragraph of the proof it was stated that $\langle \bar a \rangle \cap \overline K = \langle\bar e \rangle$ where $\bar e = \langle b \rangle.$ Why is it so?

Any help regarding this will be highly appreciated. Thanks for your valuable time.