# Cauchy's theorem on subgroup existence: intuitive version

Cauchy’s The­o­rem states that if $$G$$ is a finite group, and $$p$$ is a prime di­vid­ing the or­der of $$G$$, then $$G$$ has an el­e­ment of or­der $$p$$. Equiv­a­lently, it has a sub­group of or­der $$p$$.

This the­o­rem is very im­por­tant as a par­tial con­verse to La­grange’s the­o­rem: while La­grange’s the­o­rem gives us re­stric­tions on what sub­groups can ex­ist, Cauchy’s the­o­rem tells us that cer­tain sub­groups definitely do ex­ist. In this di­rec­tion, Cauchy’s the­o­rem is gen­er­al­ised by the Sy­low the­o­rems.

# Proof

Let $$G$$ be a group, and write $$e$$ for its iden­tity. We will try and find an el­e­ment of or­der $$p$$.

What does it mean for an el­e­ment $$x \not = e$$ to have or­der $$p$$? Noth­ing more nor less than that $$x^p = e$$. (This is true be­cause $$p$$ is prime; if $$p$$ were not prime, we would ad­di­tion­ally have to stipu­late that $$x^i$$ is not $$e$$ for any smaller $$i < p$$.)

Now fol­lows the magic step which casts the prob­lem in the “cor­rect” light.

Con­sider all pos­si­ble com­bi­na­tions of $$p$$ el­e­ments from the group. For ex­am­ple, if $$p=5$$ and the group el­e­ments are $$\{ a, b, c, d, e\}$$ with $$e$$ the iden­tity, then $$(e, e, a, b, a)$$ and $$(e,a,b,a,e)$$ are two differ­ent such com­bi­na­tions. Then $$x \not = e$$ has or­der $$p$$ if and only if the com­bi­na­tion $$(x, x, \dots, x)$$ mul­ti­plies out to give $$e$$.

The rest of the proof will be sim­ply fol­low­ing what this magic step has told us to do.

Since we want to find some $$x$$ where $$(x, x, \dots, x)$$ mul­ti­plies to give $$e$$, it makes sense only to con­sider the com­bi­na­tions which mul­ti­ply out to give $$e$$ in the first place. So, for in­stance, we will ex­clude the tu­ple $$(e, e, a, b, a)$$ from con­sid­er­a­tion if it is the case that $$eeaba$$ is the iden­tity (equiv­a­lently, that $$aba = e$$). For con­ve­nience, we will la­bel the set of all the al­lowed com­bi­na­tions: let us call it $$X$$.

Of the tu­ples in $$X$$, we are look­ing for one with a very spe­cific prop­erty: ev­ery en­try is equal. For ex­am­ple, the tu­ple $$(a,b,c,b,b)$$ is no use to us, be­cause it doesn’t tell us an el­e­ment $$x$$ such that $$x^p = e$$; it only tells us that $$abcbb = e$$. So we will be done if we can find a tu­ple in $$X$$ which has ev­ery el­e­ment equal (and which is not the iden­tity).

We don’t re­ally have any­thing to go on here, but it will surely help to know the size of $$X$$: it is $$|G|^{p-1}$$, since ev­ery el­e­ment of $$X$$ is de­ter­mined ex­actly by its first $$p-1$$ places (af­ter which the last is fixed). There are no re­stric­tions on the first $$p-1$$ places, though: we can find a $$p$$th el­e­ment to com­plete any col­lec­tion of $$p-1$$.

If $$p=5$$, and we have the first four el­e­ments $$(a, a, b, e, \cdot)$$, then we know the fifth el­e­ment must be $$b^{-1} a^{-2}$$. In­deed, $$aabe(a^{-1} a^{-2}) = e$$, and in­verse el­e­ments of a group are unique, so noth­ing else can fill the last slot.

So we have $$X$$, of size $$|G|^{p-1}$$; we have that $$p$$ di­vides $$|G|$$ (and so it di­vides $$|X|$$); and we also have an el­e­ment $$(e,e,\dots,e)$$ of $$X$$. But no­tice that if $$(a_1, a_2, \dots, a_p)$$ is in $$X$$, then so is $$(a_2, a_3, \dots, a_p, a_1)$$; and so on through all the ro­ta­tions of an el­e­ment. We can ac­tu­ally group up the el­e­ments of $$X$$ into buck­ets, where two tu­ples are in the same bucket if (and only if) one can be ro­tated to make the other.

How big is a bucket? If a bucket con­tains some­thing of the form $$(a, a, \dots, a)$$, then of course that tu­ple can only be ro­tated to give one thing, so the bucket is of size $$1$$. But if the bucket con­tains any tu­ple $$T$$ which is not “the same el­e­ment re­peated $$p$$ times”, then there must be ex­actly $$p$$ things in the bucket: namely the $$p$$ things we get by ro­tat­ing $$T$$. (Ex­er­cise: ver­ify this!)

The bucket cer­tainly does con­tain ev­ery ro­ta­tion of $$T$$: we defined the bucket such that if a tu­ple is in the bucket, then so are all its ro­ta­tions. More­over, ev­ery­thing in the bucket is a ro­ta­tion of $$T$$, be­cause two tu­ples are in the bucket if and only if we can ro­tate one to get the other; equiv­a­lently, $$A$$ is in the bucket if and only if we can ro­tate it to get $$T$$.

How many such ro­ta­tions are there? We claimed that there were $$p$$ of them. In­deed, the ro­ta­tions are pre­cisely $$(a_1, a_2, \dots, a_p), (a_2, a_3, \dots, a_p, a_1), \dots, (a_{p-1}, a_p, a_1, \dots, a_{p-2}), (a_p, a_1, a_2, \dots, a_{p-1})$$\$ and there are $$p$$ of those; are they all dis­tinct? Yes, they are, and this is be­cause $$p$$ is prime (it fails when $$p=8$$, for in­stance, be­cause $$(1,1,2,2,1,1,2,2)$$ can be ro­tated four places to give it­self). In­deed, if we could ro­tate the tu­ple $$T$$ non­triv­ially (that is, strictly be­tween $$1$$ and $$p$$ times) to get it­self, then the in­te­ger $$n$$, the “num­ber of places we ro­tated”, must di­vide the in­te­ger “length of the tu­ple”. ex­plain this bet­ter But that tells us that $$n$$ di­vides the prime $$p$$, so $$n$$ is $$1$$ or $$p$$ it­self, and so ei­ther the tu­ple is ac­tu­ally “the same el­e­ment re­peated $$p$$ times” (in the case $$n=1$$) <div><div>note: But we’re only con­sid­er­ing $$T$$ which is not of this form!%%, or the ro­ta­tion was the “stupid” ro­ta­tion by $$p$$ places (in the case $$n=p$$). %%

OK, so our buck­ets are of size $$p$$ ex­actly, or size $$1$$ ex­actly. We’re di­vid­ing up the mem­bers of $$X$$ - that is, $$|G|^{p-1}$$ things—into buck­ets of these size; and we already have one bucket of size $$1$$, namely $$(e,e,\dots,e)$$. But if there were no more buck­ets of size $$1$$, then we would have $$p$$ di­vid­ing $$|G|^{p-1}$$ and also di­vid­ing the size of all but one of the buck­ets. Mix­ing to­gether all the other buck­ets to ob­tain an uber-bucket of size $$|G|^{p-1} - 1$$, the to­tal must be di­visi­ble by $$p$$, since each in­di­vi­d­ual bucket was noteCom­pare with the fact that the sum of even num­bers is always even; that is the case that $$p=2$$.; so $$|G|^{p-1} - 1$$ is di­visi­ble by $$p$$. But that is a con­tra­dic­tion, since $$|G|^{p-1}$$ is also di­visi­ble by $$p$$.

So there is an­other bucket of size $$1$$. A bucket of size $$1$$ con­tains some­thing of the form $$(a,a,\dots,a)$$: that is, it has shown us an el­e­ment of or­der $$p$$.

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