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 Jun 5 comment prove that , there is no element $a , b$ of the group $G$ which satisfy this property @MathsLover: You must be careful when considering quotient groups. For example, if $G=\mathbb Z$ and $H=2\mathbb Z$ is a normal subgroup of $G$, then $G/H$ is isomorphic to $\mathbb Z_2$. Note that although $H$ is isomorphic to $G$, you cannot conclude that $G/H$ is isomorphic to $G/G$. By the way, the notation $(\mathbb Z_8\times \mathbb Z_4)/(\mathbb Z_4\times \mathbb Z_2)$ in my last comment is not rigorous, because $\mathbb Z_4\times \mathbb Z_2$ is even not a subgroup of $\mathbb Z_8\times \mathbb Z_4$. What I exactly meant was $((x)\times(y))/((x^2)\times(y^2))$. Jun 5 comment prove that , there is no element $a , b$ of the group $G$ which satisfy this property @MathsLover: However, since $|G|=32$ and since $|H|=8$, $G/H$ is a group of order $4$. You can check that $G/H$ is generated by a single element $xH$, i.e. $G/H$ is a cyclic group of order $4$. Jun 5 comment prove that , there is no element $a , b$ of the group $G$ which satisfy this property @MathsLover: For $G=(x)\times(y)$, $H$ is only isomorphic to $\mathbb Z_4\times \mathbb Z_2$, but not equal to $(x^2)\times (y^2)$, so you cannot obtain an isomorphism between $G/H$ and $(\mathbb Z_8\times \mathbb Z_4)/(\mathbb Z_4\times \mathbb Z_2)$. Jun 5 comment Prove that the circle $S^1$ is not the boundary of any compact manifold with boundary in $\mathbb R^2-{(0,0)}$ @tagb78: For a $1$-form $w=pdx+qdy$, by definition, $dw=(\frac{\partial q}{\partial x}-\frac{\partial p}{\partial y})dx\wedge dy$. Jun 5 comment Analytic continuation on the unit disc You are welcome! Actually there are more pathological examples: $f$ could be not analytic at any point in $\partial D$. I hope you or someone else would provide an answer with more elaborate discussion. Jun 5 comment Analytic continuation on the unit disc Here is a counter-example. Jun 5 comment A boundary version of Cauchy's theorem @MalikYounsi: I agree with you, but I still look forward to a more elementary proof. Jun 5 answered prove that , there is no element $a , b$ of the group $G$ which satisfy this property Jun 4 comment Given $\Sigma a_n$ diverges show that $\Sigma \frac{a_n}{1+a_n}$ diverges. Please note that this is not a duplicate of the linked question, because here $a_n$ is not assumed to be positive. Jun 4 comment Given $\Sigma a_n$ diverges show that $\Sigma \frac{a_n}{1+a_n}$ diverges. @Potato: Thank you! Jun 4 answered Given $\Sigma a_n$ diverges show that $\Sigma \frac{a_n}{1+a_n}$ diverges. Jun 4 comment Correspondence between modes of convergence and metrics When $\mu$ is $\sigma$-finite, it seems that you may make use of the given $\rho$ to construct a similar metric. Jun 4 comment $f_{n+1}(x):= \int_a ^x f_n(t)dt$, $\sum_{m=1} ^{\infty} f_m(x)$ is uniformly convergent @Hagrid: You are welcome. I think your proof of the uniform convergence is essentially correct. Just a little correction: $b$ should be replaced by $b-a$. Jun 4 comment Relationship between two projectors Can you prove either "if" part or "only if" part by yourself? Jun 4 comment Correspondence between modes of convergence and metrics Do you assume that $\mu$ is $\sigma$-finite? Jun 4 comment $f_{n+1}(x):= \int_a ^x f_n(t)dt$, $\sum_{m=1} ^{\infty} f_m(x)$ is uniformly convergent @Hagrid: Firstly, $(1)$ implies that $(3)$ holds for $n=1$. Secondly, can you prove $(3)$ inductively on $n$ by using $(2)$ and the fact that $\int_a^x (t-a)^{n-1}dt=\frac{(x-a)^n}{n}$? Jun 4 comment Laplace equation with periodic boundary conditions I am afraid I cannot help you much more with this, but I suggest that you apply the method of separation of variables to find some explicit solutions first. Jun 4 comment Laplace equation with periodic boundary conditions Then applying this method you may find that the space of solutions is infinite dimensional. Jun 4 comment Laplace equation with periodic boundary conditions Do you know the method of separation of variables? Jun 4 comment Prove that the circle $S^1$ is not the boundary of any compact manifold with boundary in $\mathbb R^2-{(0,0)}$ If $S^1=\partial M$, then Stokes's theorem says $\int_{S^1} w=\int_M dw$, but $dw=0$ on $\mathbb R^2\setminus\{(0,0)\}$.