# On Lagrange's Theorem [duplicate]

I have a group of order $p^\alpha$, $p$ is prime. I argued that by the Lagrange's theorem, the subgroups of this group are of prime power order. Just want to clarify if this also implies the existence of these subgroups? Can I say this group has subgroups of order $p$, $p^2$, ..., $p^{\alpha-1}$? Or do I need to refer to other results/theorems? Thank you.

## marked as duplicate by Sahiba Arora, J.-E. Pin, Matthew Towers, ahulpke, GNUSupporter 8964民主女神 地下教會Jan 10 '18 at 21:29

• "the subgroups of this group are of prime order" No, $p^2$ is not a prime, yet there may be subgroups of that order. But yes, all subgroups have order some power of $p$. However, Lagrange's theorem says nothing about whether they actually exist. They do, but not because of Lagrange's. – Arthur Jan 10 '18 at 10:53
• No, Lagrange's theorem by itself does not imply the existence of those subgroups. It only tells you that if subgroups exist, then their order will be a power of $\;p\;$ . Using a little more advanced weapons, you can even prove that for any $\;k,\,\,0\le k\le \alpha\;$ , there exists a normal subgroup of that group with order $\;p^k\;$ . – DonAntonio Jan 10 '18 at 10:57
• @DonAntonio that's wrong. you don't always have a normal subgroup. – Kenny Lau Jan 10 '18 at 11:59
• @KennyLau Yes ,you do... in finite $\;p\,-$ groups, of course . The general spirit of a proof is described in Andrea's answer. OTOH, can you name a finite $\;p\,-$ group, say of order $\;p^n\;$ ,such that for some $\;0\le k\le n\;$ there is no normal subgroup of order $\;p^n \;$ ? – DonAntonio Jan 10 '18 at 13:22
• @DonAntonio never mind. it's a theorem here – Kenny Lau Jan 10 '18 at 13:42

Lagrange's theorem implies that all subgroups are of $p$-power order. It does not imply that all possible subgroup sizes are achieved, although they are.

No, Lagrange's theorem says nothing about the existence of such groups. However, because we're talking about groups of prime power order, other theorems apply. I cannot recall a route to the proof, but somehow using Sylow's Theorems, it is shown that if a group $G$ has order $p^n$, then it has subgroups of order $p,p^2, \dots, p^{n-1}$.

• Both proofs I know of Cauchy's Theorem don't use Sylow's Theorems at all (mainly, McKay's proof...). How do Sylow theorems imply Cauchy's? – DonAntonio Jan 10 '18 at 11:02
• @DonAntonio - Honestly I got that from Wikipedia. Apparently, "[using] the fact that a p-group is solvable...", Cauchy's Theorem is a weaker form of Sylow's first theorem. I've not followed or written such a proof myself though. At least not that I can recall. – Myridium Jan 10 '18 at 11:05
• @Myridium Thanks for the explanation. I thought maybe there's some hidden (to me) way to more or less directly prove Cauchy's theorem from Sylow's...And the usual way to show the existence of subgroups of any order dividing $\;p^\alpha\;$ is, as far as I know, using the powerful fact that a finite $\;p\,-$ group has non-trivial center...and then induction. – DonAntonio Jan 10 '18 at 11:08
• @DonAntonio - upon some more digging, it seems that most treatments of Sylow's Theorems do prove the result that if a prime power divides $|G|$, then there's a subgroup of that order. This is of course a more general version of Cauchy's Theorem. – Myridium Jan 10 '18 at 11:12

A group of order $p^n$ where $p$ is a prime number is called a $p$-group.

One knows if $G$ is a $p$-group then $G$ has subgroups of order any divisor, i.e. it has a subgroup with $p^k$ elements for every $k\in\{0,1,...,n\}$.

This can be proven as follows:

• first prove that the center $Z(G)$ is non-trivial (the center $Z(G)$ is the subgroup of elements in $G$ commuting with every other element). Recall that $Z(G)$ is abelian and a normal subgroup in $G$. This fact can be obtained characterizing the center as the set of fixed points of the action $g\mapsto gxg^{-1}$ of $G$ on itself.

• set up an induction argument using (1) that for finite abelian groups Lagrange is inverted, (2) the quotient group $G/Z(G)$ is again a $p$-group of order strictly smaller than that of $G$.