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.

  • $\begingroup$ 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. $\endgroup$ Nov 27, 2023 at 20:06

1 Answer 1


I'd wager that your suspicion is correct. :)

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.

And, very-seriously, you should always trust your instincts!!!! Sure, refining and improving instincts is a big part of the math biz, but reflexive self-doubt is a very bad strategy! Better to guess wrong than to be inhibited from guessing! :)

(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.)

  • $\begingroup$ 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! $\endgroup$
    – cable
    Nov 27, 2023 at 20:10
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    $\begingroup$ 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... :) $\endgroup$ Nov 27, 2023 at 20:12
  • $\begingroup$ 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. $\endgroup$
    – cable
    Nov 28, 2023 at 17:36
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    $\begingroup$ @cable Or instead $\Bbb Z/n\Bbb Z$ they write $C_n$, which is much shorter. $\endgroup$ Nov 28, 2023 at 19:24

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