# Herstein, Topics in Algebra, page 58

## Main Question or Discussion Point

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?

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HallsofIvy
Homework Helper
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?

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
Homework Helper
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.

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.

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}

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