On the meaning of "$F^2$" in Robinson's Exercise 6.1.7.

Asked Jan 28 '22 at 23:41

Active Jan 28 '22 at 23:46

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I am confused about some notation in Exercise 6.1.7 of Robinson's, "A Course in the Theory of Groups (Second Edition)".

Here is the exercise:

Exercise 6.1.7: Let $F$ be a free group of finite rank $n$. Find the rank of $F^2$ as a free group.

The Question:

What does Robinson mean by "$F^2$"?

Thoughts:

It cannot be the free product. That's covered in the next section of the book.

It cannot be the subgroup product

$$\begin{align} F^2&=FF\\ &=\{ fg\mid f,g\in F\}\\ &=\{ h\mid h\in F\}\\ &=F \end{align}$$

because that would be silly.

It cannot be the direct product. Otherwise, we would have, for example, via this well-known result,

$$\begin{align} F_2^2&=F_2\times F_2\\ &\cong\langle a,b\mid \rangle\times \langle x,y\mid \rangle\\ &\cong\langle a,b,x,y\mid ax=xa, ay=ya, bx=xb, by=yb\rangle, \end{align}$$

which is not free.

Please help :)

edited Jan 28 '22 at 23:46

asked Jan 28 '22 at 23:41

Shaun

44,997

1

Could it be the set of words with an even number of letters? – Robert Shore Jan 29 '22 at 00:17
I don't know, @RobertShore. You reminded me that $$F^2={ f^2\mid f\in F}$$ could be an option. – Shaun Jan 29 '22 at 00:26
2

@Shaun That's not a subgroup. It might be the subgroup generated by all squares $f^2$. – Derek Holt Jan 29 '22 at 08:22
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My guess is that Derek’s right. Googling finds there’s at least one place where that notation is used. – Jeremy Rickard Jan 29 '22 at 08:32
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Assuming that $F^2 = \langle f^2 : f \in F \rangle$, you need to work out the index of $F^2$ in $F$. Then you can calculate its rank using the index formula. – Derek Holt Jan 29 '22 at 08:39

On the meaning of "$F^2$" in Robinson's Exercise 6.1.7.

The Question:

Thoughts:

0 Answers0