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Let $G$ be a group with $H_1,\ldots,H_n$ normal subgroups. Define $\varphi:G\mapsto \prod_i G/H_i$ by $\varphi(x)=(xH_1,\ldots,xH_n).$ Prove:

  1. $\ker(\varphi)=\cap_i^n H_i$,
  2. If every $H_i$ has finite index in $G$, and $|G/H_i|$ and $|G/H_j|$ are relatively prime for $i\neq j$ then $\varphi$ is a surjection and $$[G:\cap_i^n H_i]=\prod_i |G/H_i|.$$

How do I prove $\varphi$ is a surjection?

Part 1 is quite easy and the last equality follows from applying the first isomorphism theorem to $\varphi$ but I can't figure out how to use the relatively prime hypothesis to prove surjectivity.

Any help would be appreciated.

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Suppose $|H|=h$ and $k$ a positive integer such that $gcd(h,k)=1$, then we can write $1=ah+bk$ for some integers $a,b$. Thus for any $x\in H$, $x=x^{ah+bk}=(x^b)^k$. That is, $x=y^k$ for some $y\in H$.

Now let $x\in G$, we will show that $(xH_1,H_2,\dots, H_n)$ is in the image of $\varphi$. Apply the above argument for $H=G/H_1$ and $k=|G/H_2|\cdots |G/H_n|$, we see that $$\phi(x)=\phi(y^k)=(y^kH_1,y^kH_2,\dots, y^kH_n)=(xH_1,H_2,\dots, H_n).$$ Note that $y^kH_2=H_2$ because $|G/H_2|$ divides $k$. So the factor $G/H_1$ is in the image of $\varphi$. Similarly for the other factors.

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  • $\begingroup$ For this question we should show for $(x_1H_1,\dots,x_nH_n)$ there is an element $x\in G$ such that $\phi(x)=(x_1H_1,\dots,x_nH_n)$ isn't it? I can't see how the $x$ related to all the $x_1,\dots,x_n$ $\endgroup$ – Alan Wang Nov 30 '16 at 11:59
  • $\begingroup$ @AlanWang The image of a homomorphism is a subgroup. In order to show surjectivity, it suffices to show that the image contains a set of generators. $\endgroup$ – Quang Hoang Nov 30 '16 at 12:10
  • $\begingroup$ Do you mean that since $(x_1H_1,H_2,\dots,H_n),\dots,(H_1,H_2,\dots,x_nH_n)$ are in the image, so $(x_1H_1,x_2H_2,\dots,x_nH_n)$ is also in the image? $\endgroup$ – Alan Wang Nov 30 '16 at 12:15
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For maps between finite sets, to conclude bijectivity it suffices to argue the sets have the same size and the map is injective. Consider $G/\bigcap_{i=1}^n H_i\to \prod_{i=1}^n G/H_i$. We prove size equivalence.

One can prove $[G:H\cap K]\le [G:H][G:K]$ a couple ways:

  • Let $H$ act transitively from the left on the coset space $HK/K$. The element $K$ has stabilizer $H\cap K$. Orbit-stabilizer tells us $[H:H\cap K]=[HK:K]$. Therefore we may conclude that $[G:H\cap K]=[G:H][H:H\cap K]=[G:H][HK:K]\le [G:H][G:K]$.
  • Let $G$ act diagonally on $G/H\times G/K$. The stabilizer of $H\times K$ is $H\cap K$, so orbit-stabilizer says the orbit of $H\times K$ has size $[G:H\cap K]$, which is bounded by $|G/H\times G/K|$, which is precisely the product $[G:H][G:K]$.

Induct to get $[G:\bigcap_{i=1}^n H_i]\le \prod_{i=1}^n [G:H_i]$.

But by transitivity, $[G:\bigcap_{i=1}^n H_i]=[G:H_j][H_j:\bigcap_{i=1}^n H_i]$ is a multiple of $[G:H_j]$ for every index $j$. Now for some elementary number theory.

If a number $m$ is divisible by $d_1,d_2,\cdots,d_n$ which are (pairwise) coprime, what can you conclude, and what happens if you also know that $m\le d_1d_2\cdots d_n$?

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