Similar Groups

Friday 22 April 2005 at 2:57 pm | In Articles | Post Comment

Students learning (finite) group theory often have to prove that 2 groups are isomorphic. They may construct a function from G to H, guided by their Cayley tables, then assume that the function is a homomorphism. Maybe they will check a few cases but don’t think it necessary to prove all |G|^2 equations hold.

They are told that isomorphic groups have the same properties and, in particular, have the same number of elements of the same order. Unfortunately, they assume the converse is true which it isn’t. But the examples they see tend to confirm the converse; they don’t often see counter-examples.

To make things easier let’s say two finite groups G and H are similar if they have the same number of elements of the same order. I suspect this is entirely non-standard terminology 8-).

The counter-example of smallest order, 16, is where G=C_2 \times C_8 and H=<a a^2="x^8=1," ax="x^5a"> which are not isomorphic (G is abelian but H isn’t) but both groups have 1 element of order 1, 3 of order 2, 4 of order 4 and 8 of order 8.

Other examples of non-isomorphic similar groups are:

  • p is an odd prime: G=C_p \times C_{p^2},\; H=<x x^{p^2}="y^p=1," x^y="x^{1+p}"></x> which have p^2-1 elements of order p and p^3-p^2 elements of order p^2
  • p,q odd primes with q \equiv 1 \bmod{p}. Let x be an element of order p and y, z have order q. Let P=&lt; x &gt; \cong C_p and Q=&lt; y &gt; \times &lt; z &gt; \cong C_q \times C_q.  G, H are the semi-direct products of Q by P with

    • G:  y^x=y^r,\ z^x=z^r where r^p \equiv 1 \bmod{q},\ r \neq 1
    H:  \ y^x=y^r,\ z^x=z^s where r^p \equiv s^p \equiv 1 \bmod{q},\ r,s \neq 1,\ r \not\equiv s \bmod{q}Then G, H are non-isomorphic groups of order pq^2 with q^2-1 elements of order q and (p-1)q^2 elements of order p. The smallest such order is 3^2.7=147

  • q an odd prime such that q \equiv 1 \bmod{4}. Let x have order 4 and y, z order q. Let P=&lt; x &gt; \cong C_4 and Q=&lt; y &gt; \times &lt; z &gt; \cong C_q \times C_q.  G, H are the semi-direct products of Q by P with

    • G:  y^x=y^r,\ z^x=z^r
    H:  y^x=y^r,\ z^x=z^{-r}Then G, H are non-isomorphic groups of order 4q^2 with q^2 elements of order 2, 2q^2 elements of order 4 and q^2-1 elements of order q. The smallest such order is 4.5^2=100.

Agnew’s Differential Equations

Sunday 5 December 2004 at 2:59 pm | In Articles | 3 Comments

When I studied differential equations, the set book was Ralph P Agnew’s Differential Equations. It had a brilliant index which seemed to contain every word in the book. Everything I ever wanted to look up was referenced in that index, unlike plenty of other textbooks.

It included a wonderful polemic about textbooks that claim that differential equations of order n have a general solution with n essential (aka arbitrary) constants:

    The promoted the view that to each differential equation of order n there corresponds an important family of solutions from which all other solutions … are obtainable by use of appropriate hocus-pocus involving envelopes and more complicated things. It was essential that this family of solutions should have a name (this is the start of the intimidation) which would immediately convince everybody that it existed and was important. With dubious regard for appropriateness of terminology, this family was called “the general solution” of the given equation.

He then says:

    It may be unclear whether this [differential equations of order n have n essential constants] is a theorem or a definition or merely a collection of words, but we are now in a realm where nearly everything is unclear. One thing, however, is clear. No meaning has been attached to the statement that a formula has n eseential constant. This gives the good old lecturer a chance to practice the art of proof by intimidation.

He then goes on to justify his remarks and includes as examples the differential equations \displaystyle \left|\frac{dy}{dx}\right|+|x|+|y|+1=0 which has no solutions and \displaystyle \left(\frac{dy}{dx}\right)^2+y^2=1 which has “vast hordes of real solutions“.

He also deals with the solution of \displaystyle \frac{dy}{dx}=ky where dividing by y (the usual method taught at A level) won’t do since it could involve division by zero (the book indexes this as Division by zero taboo).

Finally, there is the wonderful snowplough problem (or snowplow as the author is American) which says:

    One day it started snowing at a heavy and steady rate. A snowplow started out at noon, going 2 miles the first hour and 1 mile the second hour. What time did it start snowing?

He says “Our first task is to recover from the shock of being asked to solve such a problem” and goes on “we assume that the plow clears snow at a constant rate of k cubic miles per hour“.

Now it’s up to you. Click on read more below for the time it started snowing.
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