# Category Archives: Algebra

## Simple cases of Jacobson’s theorem

A celebrated theorem of Jacobson states that

Theorem. Let $R$ be a ring, not necessarily containing $1$. If, for each $a\in R$ there exists a positive integer $n$ such that $a^n=a$, then $R$ is commutative.

This is a very strong and difficult result (although not very useful in practice). However, we can obtain some special cases via elementary means.

Proposition 1. Let $R$ be a ring such that for each $a\in R$ we have $a^2=a$. Then $R$ is commutative.

Proof. Let $a,b\in R$. Then $a+b=(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b$, i.e., $ab=-ba$. Again, $a-b=(a-b)^2=a^2-ab-ba+b^2=a-ab-ba+b$, i.e., $ab=-ba+2b$. Thus $2b=0$, i.e., $b=-b$ for each $b\in R$. Thus $ab=-ba=ba$, as desired. $\square$

The next case is already considerably harder.

Proposition 2. Let $R$ be a ring such that for each $a\in R$ we have $a^3=a$. Then $R$ is commutative.

Proof. Let $a,b\in R$. Then $a+b=(a+b)^3$ shows that

$(*)\qquad\qquad\qquad a^2b+aba+ba^2+ab^2+bab+b^2a=0$,

and $a-b=(a-b)^3$ shows that

$a^2b+aba+ba^2=ab^2+bab+b^2a$.

Hence

$(**)\qquad\qquad\qquad\qquad 2(a^2b+aba+ba^2)=0$

for all $a,b\in R$.

Plugging $a=b$ into $(**)$ gives $6a=0$, i.e., $3a=-3a$ for each $a\in R$.

Plugging $b=a^2$ into $(*)$ gives $3(a^2+a)=0$, i.e., $3a^2=3a$ for each $a\in R$. Replacing $a$ by $a+b$ gives $3(ab+ba)=0$, i.e., $3(ab-ba)=0$.

Also, multiplying $(**)$ by $a$ first on the left and then on the right and then subtracting the two gives $2(ab-ba)=0$.

From the last two paragraphs we conclude that $ab-ba=0$ for all $a,b\in R$. $\square$

Corollary. Let $R$ be a ring such that for each $a\in R$ we have $a^n=a$ for some $n\le 3$. Then $R$ is commutative.

Proof. Note that if $a^n=a$ for some $n\le 3$ then $a^3=a$. Hence the result follows by Proposition 2. $\square$

Filed under Algebra

## A nice group theory result

While working on some group theory problems today a friend and I came up with the following result.

Lemma. Let $H$ be a normal subgroup of a finite group $G$ such that $\gcd(|H|,|G/H|)=1$. If the order of $g\in G$ divides $|H|$, then $g\in H$.

Proof. Let $d$ be the order of $g$. Then the order $d'$ of $gH$ in $G/H$ divides both $d$ and $|G/H|$. But $\gcd(d,|G/H|)=1$. Hence $d'=1$, i.e., $gH=H$, i.e., $g\in H$. $\square$

Corollary 1. Let $H$ be a normal subgroup of a finite group $G$ such that $\gcd(|H|,|G/H|)=1$. If $K\le G$ such that $|K|$ divides $|H|$, then $K\le H$.

Proof. Apply the lemma to the elements of $K$. $\square$

Corollary 2. Let $H$ be a normal subgroup of a finite group $G$ such that $\gcd(|H|,|G/H|)=1$. Then $H$ is the unique subgroup of $G$ of order $|H|$.

Proof. Use Corollary 1. $\square$

Here is an example of the lemma in action.

Problem. Show that $S_4$ has no normal subgroup of order $8$ or $3$.

Solution. If $H$ is a normal subgroup of $S_4$ of order $8$, then $\gcd(|H|,|S_4/H|)=1$. Hence every element of order $2$ or $4$ in $S_4$ must lie in $H$. In particular, $(1\ 2),(1\ 2\ 3\ 4)\in H$. By a result in the previous post, $H=S_4$, a contradiction.

Likewise, if $H$ is a normal subgroup of $S_4$ of order $3$, then $H$ must contain every 3-cycle; in particular, $(1\ 2\ 3),(2\ 3\ 4)\in H$. Hence $(1\ 2\ 3)(2\ 3\ 4)=(1\ 2)(3\ 4)\in H$. But this has order 2, and $2\nmid 3$, a contradiction. $\square$

More generally, we can prove the following.

Corollary 3. $S_n$ for $n\ge 4$ has no non-trivial proper normal subgroup $H$ with $\gcd(|H|,|S_n/H|)=1$.

Proof. Suppose otherwise and let $d$ divide $|H|$. Then $H$ must contain all $d$-cycles. So if $|H|$ is even then taking $d=2$ gives $H=S_n$. If $|H|$ is odd, it contains the cycles $\sigma=(1\ \cdots\ d)$ and $\rho=(d\ \cdots\ 2\ n)$ for some $3\le d. Then $\sigma\rho=(1\ 2\ n)\in H$ has order 3. So $|H|$ contains all 3-cycles, i.e., $A_n\le H$. Since $A_n\le S_n$ is maximal, either $H=A_n$ or $H=S_n$, a contradiction. $\square$

Filed under Algebra

## Generating the symmetric group

It’s a fairly well-known fact that the symmetric group $S_n$ can be generated by the transposition $(1\ 2)$ and the $n$cycle $(1\ \cdots\ n)$. One way to prove it is as follows.

1. Show that the transpositions $(a\ b)$ for $a,b\in\{1,\dots,n\}$ generate $S_n$.
2. Show that any transposition $(a\ b)$ can be obtained from $(1\ 2)$ and $(1\ \cdots\ n)$.

We also need the following key lemma the proof of which is routine.

Lemma. $\rho (a\ b)\rho^{-1}=(\rho(a)\ \rho(b))$ for any $\rho\in S_n$ and $a,b\in\{1,\dots,n\}$.

(1) is easily proven by the observation that any permutation of $1,\dots,n$ can be obtained by swapping two elements at a time. (2) is a bit more interesting.

We first use the lemma to observe that any transposition of the form $(a\ a+1)$ can be obtained from $(1\ 2)$ upon repeated conjugation by $(1\ \cdots\ n)$. Now, since $(a\ b)=(b\ a)$, WLOG let $a. Using the lemma, conjugating $(a\ a+1)$ by $(a+1\ a+2)$ gives $(a\ a+2)$, conjugating $(a\ a+2)$ by $(a+2\ a+3)$ gives $(a\ a+3)$, etc. In this way we can eventually get $(a\ b)$. So we are done by (1).

This argument shows that $S_n$ can in fact be generated by $(a\ a+1)$ and $(1\ \cdots\ n)$ for any $a$.

Now let’s consider an arbitrary transposition $\tau$ and an $n$-cycle $\sigma$ in $S_n$. By relabeling $1,\dots,n$, we can assume that $\sigma=(1\ \cdots\ n)$ and $\tau=(a\ b)$ for $a. Note that $\sigma^{-a+1}\tau\sigma^{a-1}=(1\ c)$ where $c=b-a+1$, so WLOG $\tau=(1\ c)$. Then $\sigma^k\tau\sigma^{-k}=(1+k\ c+k)$ for each $k$. In particular, taking $k=c-1$ gives $\sigma^{c-1}\tau\sigma^{-c+1}=(c\ 2c-1)=:\rho$. Then $\rho\tau\rho^{-1}=(1\ 2c-1)$. Repeating this procedure produces $(1\ c+k(c-1))$ for $k=0,1,2,\dots$. Now $\{c+k(c-1):k=0,1,\dots,n-1\}$ is a complete set of residues mod $n$ if and only if $\gcd(c-1,n)=1$, i.e., $\gcd(b-a,n)=1$. So we’ve shown that

Theorem. Let $\tau=(a\ b)$ be a transposition and $\sigma=(c_1\ \cdots\ c_n)$ be an $n$-cycle in $S_n$. Then $\tau$ and $\sigma$ generate $S_n$ if and only if $\gcd(c_b-c_a,n)=1$.

In particular, $(a\ b)$ and $(1\ \cdots\ n)$ generate $S_n$ if and only if $\gcd(b-a,n)=1$.

Corollary. $S_p$ is generated by any transposition and any $p$-cycle for $p$ prime.

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Filed under Algebra

## Jordan-Hölder theorem

I am cramming for my algebra comprehensive exam so I will probably be posting stuff like this for a while.

A chain of subgroups $\{e\}=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_n=G$ of a finite group $G$ with $H_{i+1}/H_i$ simple for all $i$ is called a composition series of $G$. (The $H_{i+1}/H_i$ are called composition factors.) The theorem in question states that every group has a composition series, and the composition factors in any two such series are unique up to reordering.

We can easily prove the first part of the theorem, that any finite group has a composition series, using the extreme principle. Let $n\ge 0$ be the largest integer for which there is a chain $\{e\}=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_n=G$ of subgroups. (Such an $n$ exists because there is trivially a chain with $n=1$, and our group is finite.) We claim that the composition factors in this case are simple. If not, WLOG $H_{i+1}/H_i$ has a non-trivial normal subgroup $\widetilde N$. By correspondence, $\widetilde N=N/H_i$ for some $H_i\triangleleft N\triangleleft H_{i+1}$. But then

$\{e\}=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_i\triangleleft N\triangleleft H_{i+1}\cdots\triangleleft H_n=G$

is a longer chain, a contradiction.

While the proof of the general theorem requires some non-trivial group theory (e.g. see here), we can easily prove the theorem for finite abelian groups using elementary number theory. For such a group, the orders of the composition factors must exactly be the prime factors—listed with multiplicity—of the order of the group. So the result follows immediately from the fundamental theorem of arithmetic.

Filed under Algebra, Number theory

## A new AM-GM inequality

I recently found the following inequalities useful:

$\displaystyle \left(x+\frac{a_1+\cdots+a_n}{n}\right)^n\ge (x+a_1)\cdots(x+a_n)\ge \left(x+\sqrt[n]{a_1\cdots a_n}\right)^n$,

for any $x,a_1,\dots,a_n\ge 0$.

To prove the left half simply apply AM-GM to the product

$\displaystyle (x+a_1)\cdots (x+a_n)\le \left(\frac{(x+a_1)+\cdots+(x+a_n)}{n}\right)^n$.

The right half follows directly from Hölder’s inequality.

More generally, one has

$\displaystyle\left(\frac{a_1+\cdots+a_n}{n}+\frac{b_1+\cdots+b_n}{n}\right)^n\ge (a_1+b_1)\cdots(a_n+b_n)\\\ge \left(\sqrt[n]{a_1\cdots a_n}+\sqrt[n]{b_1\cdots b_n}\right)^n$

for $a_1,\dots,a_n,b_1,\dots,b_n\ge 0$. The proof goes similarly.

One obtains similar inequalities for any number of vectors $a,b,c,\dots$.

As an example, for $a_i=i$ we get the inequalities

$\displaystyle \left(x+\frac{n+1}{2}\right)^n\ge (x+1)\cdots (x+n)>\left(x+\frac ne\right)^n$

using the fact that $\sqrt[n]{n!}/n\to 1/e$ from above.