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I want to prove that there is only one group of order $5$, $(\Bbb Z/5\Bbb Z, +)$.

I know that there is only one group of every prime order, however I don't know how to prove this particular question. We're just starting group theory in this class, haven't started anything like Lagrange's Theorem yet.

So far the only idea I have is to write out every single possible "group" until I get two of the same element in a row or column in each one. That seems incredibly tedious and quite frankly I wouldn't know what all the possibilities would be.

Any advice for tackling this?

Jyrki Lahtonen
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o's1234
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    Either do the tendious work or prove it in general: https://math.stackexchange.com/questions/106163/show-that-every-group-of-prime-order-is-cyclic?rq=1 – SK19 Sep 25 '19 at 19:18
  • You could do an 'easy' version of Lagrange's Theorem for just this case - as shown below. –  Sep 25 '19 at 20:00
  • I would gladly do the tedious work, I just don't know where to begin. Would I just need to list the five tables corresponding to the five possible inverses of some element a in G, and then show that four of them aren't groups? – o's1234 Sep 26 '19 at 15:40
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    In the title, when you said “remaining the names of”, did you mean “renaming the”? – J. W. Tanner Oct 02 '19 at 07:06
  • Indeed, my bad. – o's1234 Oct 02 '19 at 19:36

1 Answers1

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Let $e$ be the identity. For some non-identity element $a$, let $A=\{ e,a,a^2,a^3, ...\}$. Let $n$ be the least power of $a$ such that $a^n=e$. Then $|A|=n\ge 2.$

If $A$ contains every element of the group, then the group is cyclic. Otherwise, there is an element $b$ not in this set. We can then let $B=\{ b,ab,a^2b,a^3b,...\}$. It is straightforward to prove that $|B|=n.$

If $A$ and $B$ have an element in common then suppose that $a^ib=a^j$. Multiplying on the left by $a^{-i}$ gives $b=a^{j-i}$, a contradiction. Therefore $A$ and $B$ are disjoint.

The group has 5 elements. Repeating this process with any element not in $A$ or $B$ would give a further $n$ elements which is impossible. But it is also impossible for the group to have $2n$ elements.

The only possibility is that the group is the cyclic group of order 5.