Riccardo's solution should work, I am guessing the problem lies here:

start: Unknown job: lpd

Guess 1: It looks like the deb is trying to restart a service which isn't running and erroring out. Try installing the lpr package from the repositories and then installing the deb and see if that works.

Guess 2: It sounds like you're trying to install a deb from a website for a brother 7340 printer: This page might be a good starting point if you want to split it off into another question.

How to fix a dpkg broken by the Brother MFC-7340 deb driver

  The cyclic group of order $6$ is the group defined by the generators $a,b$ and relations $a^2=b^3=a^{-1}b^{-1}ab=e.$


For this problem I am assuming that it must be $a,b \neq e$. I will show the $a^ib^j$ are distinct for $i=0,1$ and $j=0,1,2$. If $i=0$, $b^0=e,b,b^2$ are distinct. We know order of $b$ is not $2$ as $b^3=e$ as the order of an element divides powers that make $b$ the identity, this would mean $2\mid 3$ a contradiction. For $i=1$,
$$ab^i=ab^k \rightarrow b^i=b^k \rightarrow i\equiv k \mod{3}$$
but as $0\leq i,k &lt; 3$, $i=k$ so $a,ab,ab^2$ are distinct.

So we have that $G \supseteq \{e,a,b,b^2,ab,ab^2\}.$ So
$$G=\left&lt;a,b\right&gt; =\{ a^{n_1}b^{n_2}\cdots a^{n_r}b^{n_{r+1}}\mid n_i \in \mathbb{Z}\}$$
and as $a^{-1}b^{-1}ab=e$, $ab=ba$ and so we can write $G$ as
$$G=\{a^nb^m \mid n,m \in \mathbb{Z}.$$
We then also have by $a^{-1}b^{-1}ab=e$ that
$$a^{-1}b=ba^{-1},\,a^{-1}b^{-1}=b^{-1}a^{-1},\,ab^{-1}=b^{-1}a$$
and so any
$$a^nb^m=a\cdots a\cdot b\cdots b =b^ma^n$$
as we can commute the elements to get $b^ma^n$ so $G$ is abelian. To show there are exactly $6$ elements, if we pick any $a^nb^m \in G$, we can assume that $0\leq n\leq 1$ and $0\leq m \leq 2$ and then this element is already accounted for above so
$$G=\{e,a,b,b^2,ab,ab^2\}.$$
As $G$ is a finite abelian group of order $6$, $G$ can be written uniquely up to permutation as a direct sum of $\mathbb{Z}_{p_i^{k_i}}$'s where the $p_i$ are prime but there is only one way to do this for $6=2\cdot3$ so this means
$$G\cong \mathbb{Z}_2\oplus\mathbb{Z}_3\cong\mathbb{Z}_6$$
as $\gcd(2,3)=1$.

I am just wondering of nicer ways to argue this in the case I would need to do this on an exam. Also if everything looks correct too.
