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!
Best Answer
(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.