MHB Proving Existence of $N$ for $a^N=e$ in Finite Group $G$

[Guest2]

If $G$ is a finite group, show that there exists a positive integer $N$ such that $a^N = e$ for all $a \in G$.

All I understand is that G being finite means $G = \left\{g_1, g_2, g_3, \cdots, g_n\right\}$ for some positive integer $n.$

 

[Guest2]

I found a proof online. Let me rephrase it if I understand it. Suppose $G$ has $k$ elements. Consider the following list of the elements: $g^1, g^2, \cdots, g^{k+1}$. Since $G$ has only $k$ elements, there must be repetition in our list, say when $k = l$. So we have $g^k = g^l$ and $g^k = g^l \iff g^{k}(g^{-1})^{k} = g^l (g^{-1})^{k} \iff e = g^{l-k}$. Hence we take $N = l-k$, and the proof is complete.

I would never come up with this, however! :(

 

[Deveno]

That shows that there exists such an $N$, for any given (single) $g \in G$. But you're not done yet-you need to find an $N$ that works for EVERY $g$ in $G$.

 

[Guest2]

That shows that there exists such an $N$, for any given (single) $g \in G$. But you're not done yet-you need to find an $N$ that works for EVERY $g$ in $G$.

Thank you! I thought I was done! (Giggle)

So for any $g_i$ in our list, we have shown that we have corresponding $n_i$ such that $g_i^{n_i} = e.$ So that $n_{i-1}$ works for $g_{i-1}$ (assuming $i > 1$ here). And $n_i \times n_{i-1}$ works for both $g_i$ and $g_{i-1}$ etc. Let $N = n_1 \times n_2 \times \cdots \times n_k$. Then we've $g^N = e$ for every $g \in G.$

 

[Deveno]

Could a smaller $N$ work?

 

[Guest2]

Could a smaller $N$ work?

No, I think, since have $k$ elements this is smallest $N$ that works for every one of them. Not sure, though. (Thinking)

 

[Deveno]

Suppose I have a group of 8 elements where 5 elements have order 2, and 2 elements have order 4, and the identity has order 1.

Is $8 = (1)(2)(4)$ the smallest $N$ that will work?

You're not wrong, the multiple of all the orders will work, but it's a bit inefficient.

 

[Guest2]

Could a smaller $N$ work?

So the answer to this should have been yes, $N = \text{lcm}(n_1, n_2, \cdots, n_k)$ would work?