# $\mathbb{C}_6$, Is this a typo or something I have no comprehension of (Graduate Algebra)?

I was reviewing a previous algebra qualifying exam, and I noticed the last question included something with $$\mathbb{C}_6$$, and I realize that I don't know how we would go about even approaching that. For reference, the entire question (which I still don't even know if I want to attempt) is:

Classify up to similarity all linear transformations $$T\in \text{End}(\mathbb{C}_6)$$ such that $$T^6=0$$ and $$T$$ has at most two 2-dimensional invariant subspaces.

My assumption is that we really mean $$\mathbb{C}^6$$, but I don't trust any of my instincts anymore.

• There are several possible typos. $C_6$ is the cyclic group of order $6$, $\Bbb C^6$ is the direct product of six copies of $\Bbb C$. So you can confuse it with either $C^6$ or $\Bbb C_6$. Looks like, the second case happened. Not to mention $T^6$ or $T_6$ in your post. Nov 27, 2023 at 20:06

Alternatively, conceivably, $$\mathbb C_6$$ could have been a typo for $$C_6$$ (no blackboard bold), meaning a/the cyclic group of order $$6$$. But this seems not-so-likely given the subsequent language about "invariant subspaces", which strongly suggests that... whatever it is ... it's a vector space, not a finite cyclic group.
(There is also a fancier notion of $$\mathbb C_p$$, for prime $$p$$, denoting a completion of an algebraic closure of $$\mathbb Q_p$$, ... some such thing, but $$\mathbb C_6$$ is obviously not that.)
• Thank you! I did not know anyone wrote that as $C_6$, I am only used to $\mathbb{Z}_6$. That makes a lot of sense, thanks again! Nov 27, 2023 at 20:10
• You're welcome... and/but, $\mathbb Z_n$, though a common notation for $\mathbb Z\mod n$, is in conflict with $\mathbb Z_p$ (with $p$ prime) for the $p$-adic integers... :) Nov 27, 2023 at 20:12
• I have been avoiding the $p$-adics for too long I suppose, that explains why so many others write out $\mathbb{Z}/n\mathbb{Z}$ still. Nov 28, 2023 at 17:36
• @cable Or instead $\Bbb Z/n\Bbb Z$ they write $C_n$, which is much shorter. Nov 28, 2023 at 19:24