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Let $G = \langle a,b: a^6=b^3 =e , b^{-1}a b =a^3 \rangle$

How many elements does $G$ have?

To what family of groups is $G$ isomorphic to?

Excersise solutions help

the given relations implies that $a^2=e$

$$ G \cong \mathbb Z_{6}$$

scratch work

I am trying to set $g \in G$ as $g=a^i b^j$ and then look at what happens

for powers of $b$ alone

$$\begin{aligned} b^1=b^1 \\b^2=b^2 \\b^3=e \end{aligned} $$

for power of a alone $$ \begin{aligned} a^1&=a^1 \\a^2 &=a^2 \\a^3&= b^{-1}a b \\a^4 &= a^3 * a = b^{-1}a b \\ a^5 &= a^3 a^2 = (b^-1 a b ) a^2 \\ a^6 &= (b^{-1} a b)(b^{-1}a b )= b^{-1} a b =e \Leftrightarrow a^2=e \end{aligned} $$

so there is nothing bigger thatn $a^2$ and nothing bigger than $b^3$

from there the possible elements are

$$ a^1 , b^1 , b^2 , a^1b^1, a^1b^2 ,e$$

Need to argument for $G \cong Z_6$

what I am really tring to do is to warm up to make an argument that Diclycic group has $4n$ elements. that is another question

Thomas Andrews
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Tiger Blood
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2 Answers2

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From $a^2=e$, you get that $bab^{-1}=a^3=a^2\cdot a=a,$ hence $ba=ab.$

So $$G\cong \langle a,b\mid a^2=e,b^3=e, ab=ba\rangle$$

So this group is $\mathbb Z_2\times\mathbb Z_3\cong \mathbb Z_{6}$ by the general rule:

If $\gcd(m,n)=1$ then $\mathbb Z_{m}\times\mathbb Z_{n}\cong \mathbb Z_{mn}.$

which is essentially the Chinese remainder theorem.

Or you can skip that general theorem and just prove that $ab$ is a generator. We have $$\begin{align}(ab)^1&=ab\\(ab)^2&=b^2\\(ab)^3&=a\\(ab)^4&=b\\(ab)^5&=ab^2\\(ab)^6&=e\end{align}$$

Or just use that there are two groups of order $6$, $\mathbb Z_6$ and $S_3,$ but $S_3$ is non-abelian.

Or you can find an explicit isomorphism: $G\to\mathbb Z_6$ with $a\mapsto 3$ and $b\mapsto 2.$ Since they have the same size, and this is onto, it must also be one-to-one.

Thomas Andrews
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1

$b^{-1}a b =a^3$ implies $o(a)=o(a^3)$.

  • $o(a)=6 \implies o(a^3)=2$, contradicts $o(a)=o(a^3)$

  • $o(a)=3 \implies o(a^3)=1$, contradicts $o(a)=o(a^3)$

Therefore $o(a)=1$ or $o(a)=2$.

  • $o(a)=1 \implies G = \langle b: b^3 =e\rangle \cong C_3$.

  • $o(a)=2 \implies ab=ba \implies o(ab)=6 \implies G \cong C_6$

lhf
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  • You mean "... or $\sigma(a)=2,$" right? – Thomas Andrews Mar 14 '18 at 16:44
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    The point is, this is a group presentation. This question does not have multiple answers. – Thomas Andrews Mar 14 '18 at 16:45
  • @ThomasAndrews, you're right of course. If this is a group presentation, then you can't deduce that $o(a)=1$ and so $o(a)=2$. MY reading was the $G$ was generated by $a,b$ subject to the relations given, a subtle difference. – lhf Mar 14 '18 at 19:40