# A nonabelian $$p$$-group

Consider a prime number $$p$$ and a finite p-group $$G$$, i.e. a group of order $$p^n$$ with $$n \ge 1$$.

If $$n=1$$ the group $$G$$ is cyclic hence abelian.

For $$n=2$$, $$G$$ is also abelian. This is a consequence of the fact that the center $$Z(G)$$ of a $$p$$-group is non-trivial. Indeed if $$\vert Z(G) \vert =p^2$$ then $$G=Z(G)$$ is abelian. We can’t have $$\vert Z(G) \vert =p$$. If that would be the case, the order of $$H=G / Z(G)$$ would be equal to $$p$$ and $$H$$ would be cyclic, generated by an element $$h$$. For any two elements $$g_1,g_2 \in G$$, we would be able to write $$g_1=h^{n_1} z_1$$ and $$g_2=h^{n_1} z_1$$ with $$z_1,z_2 \in Z(G)$$. Hence $g_1 g_2 = h^{n_1} z_1 h^{n_2} z_2=h^{n_1 + n_2} z_1 z_2= h^{n_2} z_2 h^{n_1} z_1=g_2 g_1,$ proving that $$g_1,g_2$$ commutes in contradiction with $$\vert Z(G) \vert < \vert G \vert$$. However, all $$p$$-groups are not abelian. For example the unitriangular matrix group $U(3,\mathbb Z_p) = \left\{ \begin{pmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1\end{pmatrix} \ | \ a,b ,c \in \mathbb Z_p \right\}$ is a $$p$$-group of order $$p^3$$. Its center $$Z(U(3,\mathbb Z_p))$$ is $Z(U(3,\mathbb Z_p)) = \left\{ \begin{pmatrix} 1 & 0 & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix} \ | \ b \in \mathbb Z_p \right\},$ which is of order $$p$$. Therefore $$U(3,\mathbb Z_p)$$ is not abelian.

# Raabe-Duhamel’s test

The Raabe-Duhamel’s test (also named Raabe’s test) is a test for the convergence of a series $\sum_{n=1}^\infty a_n$ where each term is a real or complex number. The Raabe-Duhamel’s test was developed by Swiss mathematician Joseph Ludwig Raabe.

It states that if:

$\displaystyle \lim _{n\to \infty }\left\vert{\frac {a_{n}}{a_{n+1}}}\right\vert=1 \text{ and } \lim _{{n\to \infty }} n \left(\left\vert{\frac {a_{n}}{a_{{n+1}}}}\right\vert-1 \right)=R,$
then the series will be absolutely convergent if $$R > 1$$ and divergent if $$R < 1$$. First one can notice that Raabe-Duhamel's test maybe conclusive in cases where ratio test isn't. For instance, consider a real $$\alpha$$ and the series $$u_n=\frac{1}{n^\alpha}$$. We have $\lim _{n\to \infty } \frac{u_{n+1}}{u_n} = \lim _{n\to \infty } \left(\frac{n}{n+1} \right)^\alpha = 1$ and therefore the ratio test is inconclusive. However $\frac{u_n}{u_{n+1}} = \left(\frac{n+1}{n} \right)^\alpha = 1 + \frac{\alpha}{n} + o \left(\frac{1}{n}\right)$ for $$n$$ around $$\infty$$ and $\lim _{{n\to \infty }} n \left(\frac {u_{n}}{u_{{n+1}}}-1 \right)=\alpha.$ Raabe-Duhamel's test allows to conclude that the series $$\sum u_n$$ diverges for $$\alpha <1$$ and converges for $$\alpha > 1$$ as well known.

When $$R=1$$ in the Raabe’s test, the series can be convergent or divergent. For example, the series above $$u_n=\frac{1}{n^\alpha}$$ with $$\alpha=1$$ is the harmonic series which is divergent.

On the other hand, the series $$v_n=\frac{1}{n \log^2 n}$$ is convergent as can be proved using the integral test. Namely $0 \le \frac{1}{n \log^2 n} \le \int_{n-1}^n \frac{dt}{t \log^2 t} \text{ for } n \ge 3$ and $\int_2^\infty \frac{dt}{t \log^2 t} = \left[-\frac{1}{\log t} \right]_2^\infty = \frac{1}{\log 2}$ is convergent, while $\frac{v_n}{v_{n+1}} = 1 + \frac{1}{n} +\frac{2}{n \log n} + o \left(\frac{1}{n \log n}\right)$ for $$n$$ around $$\infty$$ and therefore $$R=1$$ in the Raabe-Duhamel’s test.

# Subset of elements of finite order of a group

Consider a group $$G$$ and have a look at the question: is the subset $$S$$ of elements of finite order a subgroup of $$G$$?

The answer is positive when any two elements of $$S$$ commute. For the proof, consider $$x,y \in S$$ of order $$m,n$$ respectively. Then $\left(xy\right)^{mn} = x^{mn} y^{mn} = (x^m)^n (y^n)^m = e$ where $$e$$ is the identity element. Hence $$xy$$ is of finite order (less or equal to $$mn$$) and belong to $$S$$.

### Example of a non abelian group

In that cas, $$S$$ might not be subgroup of $$G$$. Let’s take for $$G$$ the general linear group over $$\mathbb Q$$ (the set of rational numbers) of $$2 \times 2$$ invertible matrices named $$\text{GL}_2(\mathbb Q)$$. The matrices $A = \begin{pmatrix}0&1\\1&0\end{pmatrix},\ B=\begin{pmatrix}0 & 2\\\frac{1}{2}& 0\end{pmatrix}$ are of order $$2$$. They don’t commute as $AB = \begin{pmatrix}\frac{1}{2}&0\\0&2\end{pmatrix} \neq \begin{pmatrix}2&0\\0&\frac{1}{2}\end{pmatrix}=BA.$ Finally, $$AB$$ is of infinite order and therefore doesn’t belong to $$S$$ proving that $$S$$ is not a subgroup of $$G$$.

# Counterexamples around Cauchy condensation test

According to Cauchy condensation test: for a non-negative, non-increasing sequence $$(u_n)_{n \in \mathbb N}$$ of real numbers, the series $$\sum_{n \in \mathbb N} u_n$$ converges if and only if the condensed series $$\sum_{n \in \mathbb N} 2^n u_{2^n}$$ converges.

The test doesn’t hold for any non-negative sequence. Let’s have a look at counterexamples.

### A sequence such that $$\sum_{n \in \mathbb N} u_n$$ converges and $$\sum_{n \in \mathbb N} 2^n u_{2^n}$$ diverges

Consider the sequence $u_n=\begin{cases} \frac{1}{n} & \text{ for } n \in \{2^k \ ; \ k \in \mathbb N\}\\ 0 & \text{ else} \end{cases}$ For $$n \in \mathbb N$$ we have $0 \le \sum_{k = 1}^n u_k \le \sum_{k = 1}^{2^n} u_k = \sum_{k = 1}^{n} \frac{1}{2^k} < 1,$ therefore $$\sum_{n \in \mathbb N} u_n$$ converges as its partial sums are positive and bounded above. However $\sum_{k=1}^n 2^k u_{2^k} = \sum_{k=1}^n 1 = n,$ so $$\sum_{n \in \mathbb N} 2^n u_{2^n}$$ diverges.

### A sequence such that $$\sum_{n \in \mathbb N} v_n$$ diverges and $$\sum_{n \in \mathbb N} 2^n v_{2^n}$$ converges

Consider the sequence $v_n=\begin{cases} 0 & \text{ for } n \in \{2^k \ ; \ k \in \mathbb N\}\\ \frac{1}{n} & \text{ else} \end{cases}$ We have $\sum_{k = 1}^{2^n} v_k = \sum_{k = 1}^{2^n} \frac{1}{k} – \sum_{k = 1}^{n} \frac{1}{2^k} > \sum_{k = 1}^{2^n} \frac{1}{k} -1$ which proves that the series $$\sum_{n \in \mathbb N} v_n$$ diverges as the harmonic series is divergent. However for $$n \in \mathbb N$$, $$2^n v_{2^n} = 0$$ and $$\sum_{n \in \mathbb N} 2^n v_{2^n}$$ converges.