Recommended Air Purifier for Haze

Recently the Haze in Singapore is getting worse. One can literally smell the burning smell in the air.

A recommended cheap and good air purifier for home usage is the Xiaomi Air purifier. Xiaomi is well-known for producing high-tech goods at affordable prices. For instance, their weighing scale and fitness band (Mi Band) is top-notch for their price range.

[S$169.00](▼30%)[Local Warranty] XIAOMI Air Purifier Gen 3 // 2S // Pro (Ship by 25th Sep) WWW.QOO10.SG Haze affects Wifi Speed? Recently, Singapore is plagued by the haze, arising from forest fires in Indonesia. This is a common occurrence that happens yearly. Simultaneously, I noticed a drop in Wifi speed both at home and at work. Is this a coincidence? Or could the haze particles be actually blocking/absorbing Wifi signals so that the Wifi is slowed down? What do you think? Probably the same effect occurs for smog or fog. It is well-known and established that rain affects Wifi. Hence, perhaps it is not too far-fetched to postulate that the haze affects Wifi too? Singapore Haze & Subgroup of Smallest Prime Index Recently, the Singapore Haze is getting quite bad, crossing the 200 PSI Mark on several occasions. Do consider purchasing a Air Purifier, or some N95 Masks, as the haze problem is probably staying for at least a month. Personally, I use Nasal Irrigation (Neilmed Sinus Rinse), which has tremendously helped my nose during this haze period. It can help clear out dust and mucus trapped in the nose. [S$89.90][Clean Air]2015 Air Purifier Singapore brand and 1 year warranty with HEPA Activated Carbon UV-C germicidal killer lamp silent operation and high efficiency etc

WWW.QOO10.SG

[S\$19.90]Haze Prevention~ Nasal Rinse™- Flush out mucous~germs~bacteria~and dirt internally. Clear blocked nose and very soothing.

WWW.QOO10.SG

Previously, we proved that any subgroup of index 2 is normal. It turns out that there is a generalisation of this theorem. Let $p$ be the smallest prime divisor of a group $G$. Then, any subgroup $H\leq G$ of index $p$ is normal in $G$.

Proof: Let $H$ be a subgroup of $G$ of index $p$. Let $G$ act on the left cosets of $H$ by left multiplication: $\forall x\in G$, $x\cdot gH=xgH$.

This group action induces a group homomorphism $\phi:G\to S_p$.

Let $K=\ker \phi$. If $x\in K$, then $xgH=gH$ for all $g\in G$. In particular when g=1, xH=H, i.e. $x\in H$.

Thus $K\subseteq H$. In particular, $K\trianglelefteq H$, since $\ker\phi$ is a normal subgroup of $G$.

We have $G/K\cong \phi(G)\leq S_p$. Thus $|G/K|\mid p!$.

Also note that $|G|=|G/K||K|$. Note that $|G/K|\neq 1$ since $|G/K|=[G:H][H:K]=p[H:K]\geq p$.

Let $q$ be a prime divisor of $|G/K|$. Then $q\leq p$ since $|G/K|\mid p!$. Also, $q\mid |G|$. Since $p$ is the smallest prime divisor of $|G|$, $p\leq q$. Therefore, $p=q$, i.e. $|G/K|=p$.

Then $p=p[H:K] \implies [H:K]=1$, i.e. H=K. Thus, H is normal in G.