During lectures, we got to show that $C_9$, $C_9 + 1$ are subdirectly irreducible, where
$C_n$ := $(\{0, 1, . . . , n − 1\}, (0 1 . . . n − 1))$
$C_n + 1$ := $(\{0, 1, . . . , n − 1, n\}, (0 1 . . . n − 1)(n))$
I know the usual definition of subdirect irreducibility - An algebra $A$ is subdirectly irreducible, iff for every subdirect embedding $\alpha$, there is a projection such that: $\pi_i \circ \alpha: A \rightarrow A$ is an isomorphism.
Another definition is that
An algebra is subdirectly irreducible iff the intersection of all non-trivial congruences of he algebra is also non-trivial.
However, I don´t know how to even start with proving the exercise. Sorry that I dont show any of my progress, I just struggle with even the beginning.
How would you prove that they are subdirectly irreducible and which definition would you use?
Since there has been no answer, I try my best to provide at least partial answer to my own question. I will be grateful if you add your comments or use my answer to write your own.
I am applying the \textit{Theorem 8.4} from Burris & Sankappanavar: An algebra $A$ is subdirectly irreducible iff $A$ is trivial or there is a minimum congruence in $Con A − \{\Delta\}$. In the latter case the minimum element is $\cap(Con A - \{\Delta\})$, a principal congruence, and the congruence lattice of $A$ looks like the diagram below.
The $\Delta$ denotes "diagonal congruence relation" - the congruence relation that looks like this: "$|a|b|c|d|e|f|...$" and each element is related with itself. I will use $\Delta$ for denoting such relation.
For each algebra, I will first describe their elements and operations, then find non-trivial congruence relations on them.
Then I will try to find a \textit{monolith} for each of these algebras, since an algebra is subdirectly irreducible if and only if the lattice of its congruence relations has monolith.
(A monolith $\alpha$ is a congruence of an algebra $A$ such that for any congruence $\theta \in ConA\setminus \{\Delta\}$: $\alpha \leq \theta$.)
Congruence relations on $C_9$
$\Delta$ = $|0|1|2|3|4|5|6|7|8|$.
Other relations:
I start by generating congruence relations from $0$ paired with other elements.
It is clear that $0 \theta a$ implies $1 = f(0) \theta f(a) = a + 1$ for any $a \in C_9$. Since there is odd number of elements in this algebra.
From this and the algebra operation, it is intuitive that if I try to generate a congruence relation using $0$ and $1$ or any other pair $(a, a+1)$, the cycle will eventually get me "back to the start" and all elements end up related together.
Hence, one congruence relation is of course $|012345678|$.
Then, if I try to generate anything by relating $(a, a+2)$ or $(a, a+4)$ elements, I will end up with the same congruence relation.
Therefore, I try to generate different congruence relation by $(a, a+3)$.
Assume $0 \theta 3$ for some $\theta$. This will give $1 \theta 4 \implies 2 \theta 5 \implies 3 \theta 6 \implies 4 \theta 7 \implies 5 \theta 8 \implies 6 \theta 0$. Hence, we have another congruence relation: $|036|147|258|$.
Using $(a, a+6)$ as a pair of generators will also bring this congruence relation.
Using $(a, a+7)$ or $(a, a+8)$ will again generate a congruence relation that is already listed above.
Finding the monolith
In conclusion, the (non-diagonal) congruence relations on $C_9$ are $|036147258|$ and $|036|147|258|$.
NOT SURE ABOUT THIS PART:
The $|036|147|258|$ is a monolith of the $C_9$, because if any $a \alpha b$ and $a \neq b$, $\alpha$ is an arbitrary congruence relation on $C_9$, then it follows that $0 \alpha 3 \alpha 6$.
Hence, the $C_9$ is subdirectly irreducible.
Congruence relations on $C_9+1$
$\Delta$ = $|0|1|2|3|4|5|6|7|8|9|$.
Other relations:
Similarly to the previous case, there is the congruence relation relating all elements - $|0123456789|$.
Then, what is different about $C_9 + 1$ from the previous case is the $9$ that is mapped to itself.
This immediately brings that there will be one class where the $9$ is isolated and the rest forms the congruence relation from the previous case with $C_9$. So we have $|012345678|9|$.
By similar thinking, there is also the congruence relation $|036|147|258|9|$.
Then, if I try to use the $9$ as a generator for another congruence, I fail, since for any $a \theta 9$, it would have to follow that $a + 1 \theta 9, a + 2 \theta 9... a + 8 \theta 9$. This will of course create the congruence relation $|0123456789|$ and nothing new. So we run out of all possibilities.
Finding the monolith
In conclusion, apart from the diagonal congruence relation, there are $|0123456789|$, $|012345678|9|$ and $|036|147|258|9|$.
NOT SURE ABOUT THIS PART:
*The monolith is $|036|147|258|9|$, because it is contained in each of the other congruence relations. E.g. if $a \theta b$ for any $a \neq b$, then $0 \theta 3 \theta 6$.
So the $C_9 + 1$ is also subdirectly irreducible.