Do some cyclic subgroups "partition" a group? Is this statement true?

Let $G=\{g_1,...,g_n\}$ be a finite group. Then there exists some
$g_{k_1},...,g_{k_j}$ such that $\langle
 g_{n_1}\rangle\setminus\{1\},...,\langle g_{n_j}\rangle\setminus\{1\}$
is a partition of $G\setminus \{1\}$.

For example, $S_3=\{1,x,x^2,y,xy,x^2y\}$, where $x=(1,2,3)$ and $y=(1,2)$. Then $\langle x\rangle=\{1,x,x^2\}$, $\langle y\rangle=\{1,y\}$, $\langle xy\rangle=\{1,xy\}$, $\langle x^2y\rangle=\{1,x^2y\}$ are the cyclic subgroups such that, after discarding $1$, they partition $S_3\setminus\{1\}$.
I considered a relation on $G\setminus\{ 1\}$: $a\sim b$ if $a,b\in\langle g_i\rangle$ for some $g_i\in G$. The statement is true if it is an equivalence relation.
When trying to show it is an equivalence relation, I failed to show it is transitive. If $a\sim b$ and $b\sim c$, then $a=x^p,b=x^q$ for some $x\in G,p,q\in\mathbb{Z}$ and$b=y^s,c=y^t$ for some $y\in G,s,t\in\mathbb{Z}$. I guess $a,c$ are in the cyclic subgroup generated by an element of the form $x^uy^o$ but I cannot certify integers $u,o$.
If it turns out that this is false, then what is a counterexample?
 A: This was answered well in the comments, so I'm summarizing here.
Hankai Zeng, the original poster, observed that $G=Z_4\times Z_2$ is a counterexample.
$\def\elt#1#2{(#1, #2)}$
To see this, note that the putative partition into cyclic groups must include a subgroup $S$ that contains $\elt10$, and the same subgroup must also include the element $\elt20$. But similarly, the subgroup $S'$ that contains $\elt11$ must also include $\elt20$, so $S$ and $S'$ intersect nontrivially, which is forbidden unless $S=S'$.  But $\langle \elt10, \elt11\rangle$ generates all of $G$, so if $S=S'$ there is only one part in the partition.  $G$ is not cyclic, so the partition fails to be a partition into cyclic groups.
Similarly, the quaternion group $Q_8$, also of order 8, is a counterexample.  It has $$Q_8 = \{1, -1, i, -i, j, -j, k, -k\}$$ where $$i^2=j^2=k^2 = ijk = -1$$ and $$(-1)^2 = 1.$$  Some part of the partition must contain $i$, and this part must also contain $-1$.  Similarly whatever part contains $j$ must also contain $-1$.  This shows that $i$ and $j$ must be in the same part of the partition.  But $i$ and $j$ together generate $Q_8$ so the partition is trivial.
