Group presentations and subgroups

Question

How to prove that $G=\langle a,b,c\mid a^2 = b^2 = c^3 = 1, ab = ba, cac^{-1} = b, cbc^{-1} =ab\rangle$ has no subgroup of order $6$ without finding $G$?

$\bf Edit$: Given that $|G|=12$.

Answer 1

Given that $$G =\langle a,b,c\mid a^2=b^2=c^3=1,ab=ba,cac^{-1}=b,cbc^{-1}=ab\rangle$$ we will show that $G$ has no subgroup of order $6$.

Let $N$ be the subgroup of $G$ generated by $a$ and $b$. Since $a$ and $b$ commute, every element of $N$ takes the form $a^ib^j$, for $i,j\in\mathbb Z$. Since $a^2=1$ and $b^2=1$, it follows that $$ N = \{1,a,b,ab\}.$$ Now, we will show that $N\lhd G$. Let $g\in G$, and note that $g$ is a product of elements of $N$ and elements of the form $c^i$ for $i\in\mathbb Z$. Thus, $gNg^{-1}=N$ provided that $cNc^{-1}=N$. Since $cac^{-1}=b$ and $cbc^{-1}=ab$ are true by the presentation of $G$, and $a$ and $b$ generate $N$, we get $cNc^{-1}=N$. So, $N\lhd G$.

Then, the cosets of $N$ in $G$ are $N,cN,c^2N$. Because the cosets of a subgroup of a group partition the group, $|G|=|N|$ or $|G|=3|N|$, depending on whether $c\in N$ or $c\not\in N$. If $|G|=|N|$, then $G$ has no subsets with $6$ elements, so $G$ does not have a subgroup of order $6$ in this case.

Suppose, now, $|G|=3|N|$. Let $H$ be a subgroup of $G$ such that $|H|=6$. Since the index of $H$ in $G$ is at most $2$, it follows that $H\lhd G$. Thus, $H\cap N$ is a normal subgroup of $G$. By Lagrange's Theorem, $|H\cap N|$ divides both $|H|$ and $|N|$, so $|H\cap N|=1$ or $|H\cap N|=2$. If $|H\cap N|=1$, then $|HN|=|H||N|=6|N|$ despite the fact that $HN\subseteq G$. This is a contradiction. So, $|H\cap N|=2$.

We will show $a=b$. Let $x$ be the non-identity element of $H\cap N$. Since $H\cap N\lhd G$, it follows that $gxg^{-1}=x$ for all $g\in G$. We note that $x\in N$ and $x\ne 1$, so $x=a$, $x=b$ or $x=ab$. If $x=a$, then from $cac^{-1}=b$ we get $a=b$. If $x=b$, then from $cbc^{-1}=ab$, we get $ab=a$ and thus $b=1$. This is a contradiction, since we assumed $x\ne 1$. If $x=ab$, then from $$cabc^{-1}=cac^{-1}cbc^{-1}=bab=a$$ we get $ab=a$ and thus $b=1$. Then, by the presentation for $G$, $a=ab=cbc^{-1}=1$. Thus, $x=1$, a contradiction. Therefore, $a=b$ holds.

Using the presentation for $G$ and $a=b$, we have $$cac^{-1}=cbc^{-1} =ab = a^2=1$$ Thus, $a=c^{-1}c=1$, which entails $N=\{1\}$. Therefore, $|G|=3|N|=3$. This contradicts the hypothesis that $H$ is a subgroup of $G$ such that $|H|=6$.

Therefore, $G$ has no subgroups of order $6$.