$\begingroup$

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 ?

$\endgroup$ 12

2 Answers

$\begingroup$

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?

$\endgroup$ 2 $\begingroup$

Hint: $n\Bbb{Z} = \{nm : m \in \Bbb Z\}$.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy