Please tell me what the author wants to say. ("Topics in Algebra 2nd Edition" by I N. Herstein)

Question

I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
The following sentences are in this book.
I cannot understand what the author wants to say:

Let $G$ be a cyclic group of order $7$, that is, $G$ consists of all $a^i$, where we assume $a^7=e$. The mapping $\phi:a^i\to a^{2i}$, as can be ckecked trivially, is an automorphism of $G$ of order $3$, that is $\phi^3=I$. Let $x$ be a symbol which we formally subject to the following conditions: $x^3=e,x^{-1}a^ix=\phi(a^i)=a^{2i}$, and consider all formal symbols $x^ia^j$, where $i=0,1,2$ and $j=0,1,2,\dots,6$. We declare that $x^ia^j=x^ka^l$ if and only if $i\equiv k\mod 3$ and $j\equiv l\mod 7$. We multiply these symbols using the rules $x^3=a^7=e,x^{-1}ax=a^2$. For instance $(xa)(xa^2)=x(ax)a^2=x(xa^2)a^2=x^2a^4$. The reader can verify that one obtains, in this way, a non-abelian group of order $21$.

What is the symbol $x$?
From which algebraic system did this symbol $x$ come?
What is the product of $x$ and $a$?
Please tell me what the author wants to say.

Answer 1

Yes, although correct, this passage in the book is confusing. I remember being annoyed by this example when I was reading the book for the first time.

What Herstein shows is a special case of a certain construction of groups. So with a group $G$ and a suitable automorphism $\phi$ of $G$, you can construct a new group $H = \langle G, x \rangle$, where $G \trianglelefteq H$ and conjugation by $x$ acts on $G$ like $\phi$ does.

At this point in the book it would be difficult/cumbersome for Herstein to give all the necessary context and details for this construction. I would not worry about it too much and just keep reading. Honestly if someone is using the book to learn about groups for the first time, I think the way this example is presented is not so helpful.

The symbol $x$ is just a ''formal symbol''. It does not come from any algebraic system, you just pick something outside of $G$ then call it $x$. Then you define a multiplication on the set of symbols $\{x^i a^j\}$, subject to the rules $x^3 = a^7 = e, x^{-1}ax = a^2$. (Then the product of $x$ and $a$ is just $xa$, there is no other way to describe it.)

This is missing some details of course. Why would this multiplication be well defined? Why is it associative? What are the inverses?

Some context:

• More generally, suppose you have a group $G$ and an automorphism $\phi$ of $G$. Then with the same construction, you can construct a group $H = \langle G, x \rangle$, where $x$ is a formal symbol and we define multiplication such that $x^{-1}gx = \phi(g)$ for all $g \in G$. In this case $H = \{gx^i : g \in G, i \in \mathbb{Z}\}$. Moreover we construct $H$ such that $\langle x \rangle \cap G = \{e\}$.
• This construction is a semidirect product $G \rtimes \langle \phi \rangle$, which is a subgroup of the holomorph $G \rtimes \operatorname{Aut}(G)$. For more details look up terms like ''semidirect product'', ''external semidirect product'', ''holomorph'' in a textbook.
• Alternatively what Herstein constructs is the following group with generators and relations: $$\langle a,x : x^3 = a^7 = 1, x^{-1}ax = a^2 \rangle.$$ Look up ''free groups'', ''generators and relations'' in a textbook.

Answer 2

I hate descriptions of a group like that in Herstein, because it is not at all obvious to a newcomer that there might not be some secret collapsing of everything into the identity element. How do you know you can just "declare" when two things are equal without creating some kind of inconsistency somewhere?

What Herstein is aiming to do looks like describing a group by generators and relations, but typically the most you can get from such a description alone is an upper bound on the size of the group, but to know the description is really a group of order $21$ you need to put in some work. To understand what I'm getting at, the post here gives a description of a group with 3 generators and a few relations that turns out to be the trivial group after some fairly nontrivial work.

Instead of trying to make sense of what Herstein is doing, let's instead just build ourselves a nonabelian group of order $21$ using $2 \times 2$ matrices with entries in $\mathbf Z/7\mathbf Z$.

In the group $(\mathbf Z/7\mathbf Z)^\times$, the subgroup of order $3$ is $\{1,2,4 \bmod 7\} = \{b \bmod 7 : b^3 \equiv 1 \bmod 7\}$. Let $G = \{(\begin{smallmatrix}b&c\\0&1\end{smallmatrix}) \bmod 7 : b^3 = 1 \}$. The upper right entries of matrices in $G$ are arbitrary integers modulo $7$ while the upper left entries are restricted to be the elements of $(\mathbf Z/7\mathbf Z)^\times$ with order dividing $3$, and those are the powers of $2 \bmod 7$. As a set, $G$ has size $21$: $3$ options for the upper left entry and $7$ options for the upper right entry, and they can be picked arbitrarily.

Check $G$ is a closed under matix multiplication and inversion, so $G$ is a group of order $21$. In $G$, let $a = (\begin{smallmatrix}1&1\\0&1\end{smallmatrix})$ and $x = (\begin{smallmatrix}2&0\\0&1\end{smallmatrix})$. Then $a$ has order $7$, $x$ has order $3$, and $xax^{-1} = (\begin{smallmatrix}1&2\\0&1\end{smallmatrix}) = a^2$. In particular, $a$ and $x$ do not commute (since $xax^{-1} = a^2$ and $a^2 \not= a$), so $G$ is nonabelian. A general element of $G$ looks like $(\begin{smallmatrix}2^i&j\\0&1\end{smallmatrix}) = (\begin{smallmatrix}1&j\\0&1\end{smallmatrix}) (\begin{smallmatrix}2^i&0\\0&1\end{smallmatrix}) = (\begin{smallmatrix}1&1\\0&1\end{smallmatrix})^j (\begin{smallmatrix}2&0\\0&1\end{smallmatrix})^i = a^jx^i. $ Herstein writes elements of his $G$ in the form $x^ia^j$. As $i$ and $j$ vary, the set of all $a^jx^i$ equals the set of all $x^ia^j$ since every element of $G$ is the inverse of something in $G$ and $(a^jx^i)^{-1} = x^{-i}a^{-j}$.

A similar construction gives you a nonabelian group of order $pq$ where $p$ and $q$ are primes such that $q \equiv 1 \bmod p$. See the top of page 5 here. You are looking at the case $p = 3$ and $q = 7$.