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)


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


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


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.