Herstein, Topics in Algebra, page 58

Jimmy Snyder

I have the second printing of the first edition of Herstein's 'Topics in Algebra', published 1964.

On page 58 near the middle of the page there is a paragraph that begins:

Let G be a cyclic group ...

The author writes
[tex]\phi:a^i \rightarrow a^{2i}[/tex]

and later

[tex]x^{-1}a^ix = \phi(a)^i = a^{3i}[/tex]

The next paragraph makes it clear that he means:
[tex]x^{-1}a^ix = \phi^i(a) = a^{3i}[/tex]

But it doesn't seem true to me. for instance if i = 1, then no matter how I write it, I get:
[tex]\phi(a) = a^3[/tex]

but by the definition of phi,
[tex]\phi(a) = a^2[/tex]

What gives?

HallsofIvy

What is x in [tex]x^{-1}a^ix = \phi(a)^i = a^{3i}[/tex]? Surely it can't be just any member of G because then we would have a= a³.

And what is a? Any member of G or specifically a generator of G?

Jimmy Snyder

Sorry, I didn't put enough information for anyone that doesn't have a copy of the book. G is a cyclic group of order 7, a is an element of G, so that [itex]G = \{e = a^0, a^1, a^2, a^3, a^4, a^5, a^6\}[/itex]. x is a formal symbol. The author intends to describe the group of order 21 made of formal symbols [itex]x^ia^j, i = 0, 1, 2 j = 0, 1, 2, 3, 4, 5, 6[/itex].

morphism

I think it's a typo and should be [itex]a^{2i}[/itex] instead. Unfortunately I don't have my copy of Herstein on me right now to verify this. Maybe you could post a bit more of that page?

His intent is clear though: he's trying to define the semidirect product of G and X={1, x, x^2}, with X viewed as the cyclic group of order 3, where conjugation by x acts as [itex]\phi[/itex] on G.

Jimmy Snyder

Perhaps. However, he gives a specific example of multiplication in the larger group.
[tex]x^1a^1 \cdot x^1a^2 = x^1(a^1x^1)a^2 = x^1(x^1a^3)a^2 = x^2a^5[/tex]
That's taking a typo pretty far, but I suppose it's possible he lost track half way through the paragraph.

peter_n

It was a typo in the 1st edition.
The corrected expression is on pg 69 of the 2nd edition:

x^{-1}a^ix = \\phi(a^i) = a^{2i}

peter_n

With formatting....
[tex]
x^{-1}a^ix = \phi(a^i) = a^{2i}
[tex]