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I have a problem understanding the assumptions for the proof of this theorem

Theorem: If $A$ is abelian with order $a$ divisible by prime $p$, then $A$ has an element of order $p$.

The proof goes as follows: Obviously true if $|A|=p$. We may therefore employ induction and shall henceforth assume that $a$ is composite divisible by $p$. By Chapter II, Theorem 5, p. 40, $A$ possesses proper subgroups. Let us select a proper subgroups $H$ of maximum order $h$, ($h<a$) say. We have to distinguish two cases:

(i) $p \mid h$. By induction, $H$ contains an element of order $p$

(ii) $(h, p) =1$. Since $H$ is proper, there exists an element $x$ or order $t$ which does not belong to $H$. Let $T = \langle x\rangle$ and consider the product $HT$. Since $A$ is Abelian, $HT = TH$ so $HT$ is a subgroup more comprehensive than $H$ so it must be $A$ since $H$ was maximal. By the product theorem $ad = ht$ where $d$ is the order of the intersection of $H$ and $T$ which must be $\{e\}$ and therefore $d =1$ and so $a = ht$. Since $p \mid a$ and $(h, p) = 1$, then $p \mid t$ so $t = sp$ for some $s$. Then $x^s$ is of order $p$. QED.


I paraphrased only slightly but the proof is what was presented. I have a problem with the induction hypothesis and the assumption about composite order. I assume that the induction hypotheses is that “all groups whose order is divisible by $p$ and smaller than the order of $A$ have an element divisible by $p$”, but I am not quite sure.

The second problem I have is the assumption that we can restrict ourselves to composite order. Doesn’t that eliminate all groups of prime power order? I must be missing something.

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1 Answer 1

Your first assumption is correct. We are indeed using (strong) induction on the number $\frac{|A|}{p}$ for fixed, but arbitrary $p$.

Any non-identity element $a$ in any group $A$ of order a prime $p$ is itself of order $p$. That is to say, for each prime $p$ there is exactly one group (even without assuming abelian) of order $p$ and that group is cyclic. Therefore the pure prime ordered groups are dealt with in just a side comment of the proof.

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