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I think I finally have a correct answer for arbitrary $p$. As F. Ladisch notes, $G=C_{p^3}$ has only finitely many indecomposable modular representations. For the following argument, I will not only n …

If $$ \Theta\colon\quad 0\to A\to B\to C\to 0 $$ is an exact sequence of abelian groups, and $n$ is an integer, then one obtains an exact sequence $$ 0\to A[n] \to B[n] \to C[n] \stackrel{\delta_n(\Th …

I am going to assume that by "algebra" you simply mean a ring. The answer is "no", in general. For example $\mathbb{Z}/5\mathbb{Z}$ is not the unit group of a ring. Indeed, suppose it was the unit gro …

Let me try this. I think that the answer this time is positive. Step 1: We will first reduce to $p$ power order. Let $M$, $M'$ be matrices over $\mathbb{F}_p$ of order $p^na$, where $p\nmid a$. Then, …

To put this one to rest, I will answer the more precise question that, after much prodding, we got Adam to formulate in the comments. I am merely paraphrasing a comment of Qiaochu. If you are given t …

Take any polynomial of degree $e$ that is Eisenstein at $p$, adjoin to $\mathbb{Q}_p$ a root of that polynomial and you will get a totally ramified extension of $\mathbb{Q}_p$ of degree $e$. Moreover, …