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Please could someone help me verify this short proof of Lagrange's theorem? I just came up with it but am not convinced it works. I have starred the steps I'm most unsure about.

Lemma: Let $|G| = d$. If $g^a = e, a \leq d$ for some $g$, then $d = aj, j \in \mathbb{Z}$.

Proof:

Suppose $a$ is the smallest positive number such that $g^a = e$. [*]

$a \leq d$, so we can write $d = aj + r, j \in \mathbb{Z}, 0 \leq r < a$.

Then we have $g^d = g^{aj + r} = (g^a)^j \times g^r = e^j \times g^r = g^r$, so $g^r = g^d = e$. But $r < a$ and we assumed $a$ is the smallest such number, so $r = 0$. QED

Theorem (Lagrange): the order of a subgroup $H \subset G$ divides the order of the group $G$.

Proof: The lemma establishes that the order of any element divides the order of the whole group. So to complete the proof, simply notice that (a) the order of a subgroup is the maximum over the order of its elements and (b) the order of an element in a subgroup $H$ is the same as its original order in the group $G$ [**]. Therefore, the order of $H$ will be the order of some $g \in G$, which divides $d$ by the lemma. QED?

[* can I just assume this? If $a$ isn't the least such number, can't I just start again with the one that is?]

[**(b) seems intuitively clear but I'm not sure how to justify it]

Thanks for your help!

Edwin Agnew
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    To prove lagrange theorem you can do it by contradiction, assuming that there exists a subgroups whose order does not divide the order of $G$ and finding an element that does not satsify your lemma. – Marcos Mar 24 '22 at 21:41
  • Could you help me with the details of that? I'm guessing something like: suppose $|H| = d_H, gcd(d_H, d) = 1$. Then by the lemma, $h^a = e \in H \implies d_H = aj$ in the subgroup, while $h^a = e \in G \implies d = aj'$ in the group. But then $a$ divides both $d$ and $d_H$, which is a contradiction. Does this look OK? – Edwin Agnew Mar 24 '22 at 21:56
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    In the proof of the lemma, how do you know that $g^d=e$ (without presupposing Lagrange's theorem)? – Andreas Blass Mar 24 '22 at 23:31
  • Related to https://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory – lhf Mar 24 '22 at 23:48
  • Good point! I had in mind a proof that didn't use Lagrange's theorem but was only for abelian groups. Judging from that link, it does seem possible? – Edwin Agnew Mar 25 '22 at 18:11

1 Answers1

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(a) it's wrong. You can't always find an element in $H$ with order equal to the order of $H$, unless when $H$ is a cyclic group.

For example, the order of $\mathbb{Z}_2\times \mathbb{Z}_2$ is 4, but the maximum order of its elements is 2.

Compacto
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  • Nice counter-example - many thanks. Does your first sentence go in both directions? i.e. an element's order is equal to the group order iff its a cyclic group? – Edwin Agnew Mar 24 '22 at 21:48
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    Yes it does, because the order of an element $g$ of $G$ equals the order of the subgroup it generates $\langle g\rangle$. Since $\langle g\rangle$ is always a subgroup of $G$, if the order of $g$ equals the order of $G$, then $\langle g\rangle$ has the same number of elements as $G$, and so it must be $\langle g\rangle = G$. – Compacto Mar 24 '22 at 21:50