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.


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 \(p\).