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
\phi:a^i \rightarrow a^{2i}

and later

x^{-1}a^ix = \phi(a)^i = a^{3i}

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

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

but by the definition of phi,
\phi(a) = a^2

What gives?

 

[HallsofIvy]

What is x in x^{-1}a^ix = \phi(a)^i = a^{3i}? 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]

What is x in x^{-1}a^ix = \phi(a)^i = a^{3i}? 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?

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 G = \{e = a^0, a^1, a^2, a^3, a^4, a^5, a^6\}. x is a formal symbol. The author intends to describe the group of order 21 made of formal symbols x^ia^j, i = 0, 1, 2 j = 0, 1, 2, 3, 4, 5, 6.

 

[morphism]

I think it's a typo and should be a^{2i} 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 \phi on G.

 

[Jimmy Snyder]

I think it's a typo and should be a^{2i} 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 \phi on G.

Perhaps. However, he gives a specific example of multiplication in the larger group.
x^1a^1 \cdot x^1a^2 = x^1(a^1x^1)a^2 = x^1(x^1a^3)a^2 = x^2a^5
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...
<br /> x^{-1}a^ix = \phi(a^i) = a^{2i}<br />