[Math] Showing Z has infinitely many subgroups isomorphic to Z

abstract-algebragroup-isomorphismgroup-theory

I tried to do this as

Z has elements from -$\infty$ to $\infty$.

Let's take n number of elements out to $Z$.

Suppose n= 5
And the elements are 1,2,3,4,5.

Which forms a $S_5$ or a permutation group a of 5 elements .
A\c to Cayley Theorem
"Every group is isomorphic to permutation groups "
Thus $S_5$ is isomorphic to $Z$.

As n can have value 1 to $\infty$

And thus there are infinite number of permutations groups from $S_1$ to $S_{\infty}$ and all are isomorphic to $Z$.

All comprising subgroups of $Z$.
Thus Z has infinitely many subgroups isomorphic to Z .
Is this proof okay or needs some modifications ?

Best Answer

Some facts and hints, assuming that $Z=\mathbb Z$ is the integers.

  1. $\mathbb Z$ does not have any finite non-trivial subgroups.
  2. In particular, $S_5$ is not a subgroup of the integers.
  3. The groups you are looking for are infinite and free, just like $\mathbb Z$ and unlike $S_5$.
  4. Consider the even numbers $2\mathbb Z$ -- what can you say about them?
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