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Let $G$ be a finite abelian $p$-group. Let $\{x_1,\cdots,x_r\}$ be a subset of $G$ with following property:

(1) $\langle x_1, \cdots, x_r\rangle=\langle x_1\rangle \oplus \cdots \oplus \langle x_r\rangle$.

(2) No $x_i$ is a $p$-th power in $G$. [In other words, all $x_i$'s are outside Frattini subgroup $\Phi(G)$.]

Q. Can we always extend the above set to set to $\{x_1,\cdots, x_r, x_{r+1},\cdots, x_l\}$ such that $$G=\oplus_{i=1}^l \langle x_i\rangle?$$


In the proof of fundamental theorem of finite abelian group, the set $\{x_1,\cdots, x_r\}$ has an additional property: $x_1$ is of maximum order in $G$; $x_2$ is of maximum order with property that $\langle x_2\rangle$ intersects trivially with $\langle x_1\rangle$, and so on. Then one proceeds inductively to extend above set to generating set which gives cyclic decomposition of $G$.

In above problem, I am not considering any such restriction related to orders of elements.

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No. Let $G = \langle a \rangle \oplus \langle b \rangle$ where $a$ and $b$ have orders $p$ and $p^3$, respectively, and let $H = \langle x \rangle$ with $x = ab^p$. So $H \cong C_{p^2}$ and is not a direct summand of $G$.

(I once made a mistake myself in a proof and this was essentially the counterexample.)

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