Algebraic results with easier non-commutative proofs than commutative proofs It would be nice to know of results in algebra which are relevant both in a commutative and non-commutative setting, with 'easier' proofs in the non-commutative case, or at least more or less in the same 'level' as the commutative one.
I have found the following two examples and would be glad to hear about more examples:
(1) The dimension of a vector space over a field and the rank of a free module over a division ring.
(2) Theorem 1 for $\mathbb{C}[x,y]$ and Theorem 2.11 for the first Weyl algebra (it is not so difficult to show that in the commutative result we can replace $\mathbb{C}$ by any field of characteristic zero).
In my opinion, in (1) the results have 'same level', while in (2) the non-commutative result is even 'easier' than the commutative result; do you agree with me? 
Remark: Of course, the same question can be asked also in other areas of mathematics (considering a seemingly more difficult setting, in which actually things are getting easier).
 A: The algebra of quasisymmetric functions $\operatorname{QSym}$ is free commutative. This is a rather deep result by Hazewinkel, and the proof uses Lyndon words and their shuffling properties. See Theorem 6.5.13 in Darij Grinberg and Victor Reiner, Hopf Algebras in Combinatorics, version 11 May 2018 (also available as arXiv:1409.8356v5).
The Malvenuto-Reutenauer Hopf algebra $\mathbf{FQSym}$ (a noncommutative algebra that is, in a sense, a noncommutative version of $\operatorname{QSym}$) is free. This is a fairly simple exercise, using merely the decomposition of a permutation matrix into indecomposable blocks (aka "connected permutations"). See Exercise 8.1.10 ibidem.
A: The classification of Novikov algebras with non-abelian Lie algebras is much easier than the classification of Novikov algebras with abelian Lie algebras, i.e., in the commutative case - see this article, section $3$.
A: *

*If $S$ is a set, then the free monoid $S^*$ on $S$ is cancellative: That is, if $a$, $b$ and $c$ are three elements of $S^*$ satisfying either $ac = bc$ or $ca = cb$, then $a = b$. This is almost trivial: The elements of $S^*$ are tuples of elements of $S$, and multiplication in $S^*$ is just concatenation of tuples; thus, $ac = bc$ implies $a = b$ by cutting off the last few entries, whereas $ca = cb$ implies $a = b$ by cutting off the first few entries.

*If $S$ is a set, then the free abelian monoid $\mathbb{N}S$ on $S$ is cancellative. This is still not hard to prove, but less obvious. The elements of $\mathbb{N}S$ are formal $\mathbb{N}$-linear combinations of elements of $S$ or (equivalently) finite multisets of elements of $S$. In constructive logic, this becomes complicated to deal with, since equality may not be decidable in $S$; thus, a formal $\mathbb{N}$-linear combination does not have well-defined "multiplicities": e.g., if you have an $\mathbb{N}$-linear combination $x+y$ and you don't know whether $x=y$ or not, you can't tell whether $x$ appears once or twice in this combination. Cancellativity of $\mathbb{N}S$ still holds, but the proof is trickier. (Some of the ideas from https://mathoverflow.net/questions/302516 help here.)
