Confused about cyclic goups

Question

The cyclic group $\mathbb Z_6$ has a subgroup of order $3$ so I can deduce that $\mathbb Z_3$ is a subgroup of $\mathbb Z_6$. On one hand, it seems false to me cause the group operations of the above groups are not the same. On the other hand, the subgroup of order $3$ is isomorphic to $\mathbb Z_3$. I am confused...

Answer 1

No, you cannot deduce that $\Bbb Z_3$ is a subgroup of $\Bbb Z_6$. What you can deduce is that $\Bbb Z_6$ has a subgroup which is isomorphic to $\Bbb Z_3$.

Answer 2

The set $\mathbb Z/3\mathbb Z$ is not even a subset of $\mathbb Z/6\mathbb Z$.

There are three members of $\mathbb Z/3\mathbb Z$. Explicitly, they are:

\begin{align} \{\ldots,{-7},{-4},{-1},2,5,\ldots\}&\in\mathbb Z/3\mathbb Z\\ \{\ldots,{-6},{-3},0,3,6,\ldots\}&\in\mathbb Z/3\mathbb Z\\ \{\ldots,{-5},{-2},1,4,7,\ldots\}&\in\mathbb Z/3\mathbb Z \end{align}

There are six members of $\mathbb Z/6\mathbb Z$:

\begin{align} \{\ldots,{-14},{-8},{-2},4,10,\ldots\}&\in\mathbb Z/6\mathbb Z\\ \{\ldots,{-13},{-7},{-1},5,11,\ldots\}&\in\mathbb Z/6\mathbb Z\\ \{\ldots,{-12},{-6},{0},6,12,\ldots\}&\in\mathbb Z/6\mathbb Z\\ \{\ldots,{-11},{-5},{1},7,13,\ldots\}&\in\mathbb Z/6\mathbb Z\\ \{\ldots,{-10},{-4},{2},8,14,\ldots\}&\in\mathbb Z/6\mathbb Z\\ \{\ldots,{-9},{-3},{3},9,15,\ldots\}&\in\mathbb Z/6\mathbb Z \end{align}

Note that there is no $x$ such that $x\in\mathbb Z/3\mathbb Z$ and $x\in\mathbb Z/6\mathbb Z$. The two sets have no members in common.

Since it isn't a subset, $\mathbb Z/3\mathbb Z$ is not a subgroup of $\mathbb Z/6\mathbb Z$ either.

Answer 3

$\Bbb{Z}_6$ has a subgroup $\{[0],[2],[4]\}$ which is isomorphic to $\Bbb{Z}_3$, but $\Bbb{Z}_3$ is not even a subset of $\Bbb{Z}_6$.