Tag Archives: permutation

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<n. 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

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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 ncycle (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<b. 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<b. 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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Permutation of Digits by Multiplication

Here is an expository write-up of this post.

The number N=142857 is interesting because of the following property

\begin{aligned}N&=142857\\ 2N&=285714\\ 3N&=428571\\ 4N&=571428\\ 5N&=714285\\ 6N&=857142\end{aligned}

and 7N=999999. The last equation means that

\displaystyle\frac 17=0.\overline{142857}.

So multiplying N by 1,2,3,4,5,6 simply permutes the digits of N. As elements of S_6 these permutations correspond to (using cycle notation)

\hbox{id}, (153)(264), (165432), (135)(246), (123456), (14)(25)(36)

respectively. We can read off the following from this list.

  • Since (165432) and (123456) are full-length cycles, 3 and 5 are primitive roots modulo 7.
  • 2 and 4 have order 3 modulo 7, while 6 has order 2.

In general, if p is an odd prime and b is a positive integer with order d modulo p, then 1/p will have period (b^d-1)/p of length d in base b. This follows simply from observing that if b^d-1=mp then

\displaystyle\frac 1p=\frac{m}{b^d-1}=\frac{m}{b^d}+\frac{m}{b^{2d}}+\frac{m}{b^{3d}}+\cdots.

Now let b>p be a primitive root modulo p. We shall work in base b. Then 1/p will have period L=(b^{p-1}-1)/p of length p-1. Since the periods of 1/p,2/p,\dots,(p-1)/p are L,2L,\dots,(p-1)L, all of length p-1, which are all powers of some permutation of the digits of L, and no two of which are congruent modulo b, it follows that the digits of L are all distinct. So we’ve shown the following.

Theorem. If p is an odd prime and b>p is a primitive root modulo p, then the base b expansion of 1/p has period of length p-1 with distinct digits.

Taking p=7, b=10 gives our previous example. So the prime 7 in base 10 is indeed very special, since it is the only prime less than 10 for which 10 is a primitive root.

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Symmetric Polynomials in Two Variables

Let R be a commutative ring. The fundamental theorem of symmetric polynomials says that any symmetric polynomial in R[X,Y] can be expressed uniquely as a polynomial in X+Y and XY.

Recently I was thinking about this along the following lines. Let R[X,Y]^{S_2} denote the set of all symmetric polynomials in R[X,Y]. Then the theorem above is saying that R[X,Y]^{S_2} is generated by \{X+Y,XY\} as an Ralgebra, i.e., R[X,Y]^{S_2}=R[X+Y,XY]. Being unsuccessful at utilising this I ended up with the following. (I can’t see the best way of showing that a set generates an algebra.)

Let f(X,Y)=\sum_{i,j}a_{i,j}X^iY^j\in R[X,Y]^{S_2}. Since f(X,Y)=f(Y,X), we have a_{i,j}=a_{j,i} for all i,j. Hence f(X,Y)=\sum_{i\le j}a_{i,j} (XY)^i(X^{j-i}+Y^{j-i}). So it suffices to show that X^n+Y^n can be expressed as a polynomial in X+Y and XY for each n\ge 0. But this follows easily by induction and the following identity


However, this argument doesn’t generalise immediately to more variables, and I don’t particularly like any of the proofs that I’ve found so far.

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Number of Permutations of Prime Order

In the last post we counted the number of elements of order 2 in the symmetric group S_n. In exactly the same way we can count the number of elements of order p, where p is a prime number. Let this number be denoted by N_p(n). Then

\displaystyle\begin{aligned}N_p(n)&=\sum_{k=1}^{\lfloor n/p\rfloor}\frac{(p-1)!^k}{k!}\binom np\binom{n-p}{p}\cdots\binom{n-pk+p}{p}\\ &=\sum_{k=1}^{\lfloor n/p\rfloor}\frac{n!}{k!(n-pk)!p^k}\\ &=\sum_{k=1}^{\lfloor n/p\rfloor}\binom{n}{pk}\frac{(pk)!}{k!p^k}.\end{aligned}

Here we adopt the convention \binom nm=0 if m>n. Note that

\displaystyle\frac{(pk)!}{k!p^k}=\frac{(pk)!}{p\cdot 2p\cdots kp}\equiv (p-1)!^k\equiv (-1)^k\pmod p

by Wilson’s theorem. So

\displaystyle N_p(n)\equiv\sum_{k=1}^{\lfloor n/p\rfloor}(-1)^k\binom{n}{pk}\pmod p.

If n=n_rp^r+\cdots+n_1p+n_0 and k=k_{r}p^{r-1}+\cdots+k_2p+k_1 are base p expansions, then using Lucas’ theorem one has

\displaystyle\binom{n}{pk}\equiv\binom{n_r}{k_r}\cdots\binom{n_1}{k_1}\pmod p.


\displaystyle\begin{aligned}N_p(n)&\equiv -1+\sum_{k=0}^{\lfloor n/p\rfloor}(-1)^k\binom{n}{pk}\pmod p\\ &\equiv -1+\sum_{k=0}^{\lfloor n/p\rfloor}(-1)^{k_1+\cdots+k_r}\binom{n_r}{k_r}\cdots\binom{n_1}{k_1}\pmod p\\ &\equiv -1+\prod_{j=1}^r\sum_{k_j=0}^{n_j}(-1)^{k_j}\binom{n_j}{k_j}\pmod p\\ &\equiv -1\pmod p\end{aligned}

This generalizes the result from the previous post.

Corollary. The number of subgroups of order p in S_n is congruent to 1\pmod p.

Proof. If there are k subgroups of order p in S_n, then N_p(n)=k(p-1)\equiv -1\pmod p and so k\equiv 1\pmod p. \square

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