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Corollary 4.6.8. There is a group $G$ of order $n^3$ given by $G = \{b^ic^ja^k | 0 ≤ i, j, k < n\}$, where $a$, $b$, and $c$ all have order $n$, and $b$ commutes with $c$, $a$ commutes with $c$, and $aba^{−1} = bc$. Thus,

$$(b^ic^ja^k) (b^{i'}c^{j'}a^{k'}) = b^{i+i'}c^{j+j'+ki'}a^{k+k'}.$$

a) Consider the groups constructed in Corollary 4.6.8 for $n$ a prime number $p$. When $p>2$, show that every element in this group has exponent $p$. (Why is this different from the case $n=2$?) Recall that in Problem 3 of Exercises 2.1.16 it was shown that any group of exponent $2$ must be abelian. As we see here this property is not shared by primes $p>2$.

b) Now consider the groups of Corollary 4.6.8 for general values of $n$. For which values of $n$ does the group in question have exponent $n$? If the exponent is not equal to $n$, what is it?

This is what I did:

I first tried to see if there was any pattern that I could detect by multiplying the elements $b^ic^ja^k = b^ia^kc^j$ over an over again. This is what I got:

$(bac)(bac) = b^2a^ac^{2+1}$

$(b^2a^ac^{2+1})(bac) = b^3a^3c^{4+2}$

$(b^3a^3c^{4+2})(bac) = b^4a^4c^{7+3}$

$(b^4a^4c^{7+3})(bac) = b^5a^5c^{11+4}$

But I wasn't sure if this would help me in finding the order of bca...however, the corollary tells us that $aba^{-1}=bc$, so $c=b^{-1}aba^{-1}$, right?

So I tried using that, and we have:

$(bca) = bb^{-1}aba^{-1}a = ab$. However, this doesn't really help much either.

So I was wondering if anybody could help me with this...

Thank you in advance.

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up vote 3 down vote accepted

I make it $(a^ic^jb^k)^m = a^{mi}b^{mk}c^{mj+m(m-1)ki/2}$. So the exponent is $n$ when $n$ is odd and $2n$ when $n$ is even.

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