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Suppose $G'$ is a homomorphic image of $G$.
Let $\phi:G\rightarrow G'$ be an epimorphism.
We want to show that $|G'|$ divides $|G|$.

It can be done easily by using the concept of cosets, quotient group and also First Isomorphism Theorem. But this exercise appears under the subtopic homomorphisms so those contents are not involved yet.

From what I know, there is an identity that is
$$|G|=|\ker \phi||G'|$$ But I have no idea how to prove it.
Does this mean that the epimorphism will partition image of $G$ into $G'$ classes where each class have $|\ker \phi|$ elements?

I just write my answer in a proper way. Since $\phi$ is surjective, the pre-images of every element in $G'$ will partition $G$. That is, $$|G|=\sum_{y\in G'}|\phi^{-1}(y)|$$ Let $y\in G'$
Then $y=\phi(x)$ for some $x\in G$ due to the surjectivity.
Hence, $|\phi^{-1}(y)|=|\phi^{-1}(\phi(x))|=|x\cdot \ker \phi|=|\ker \phi|$.
This proves the identity.

Wang Kah Lun
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2 Answers2

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It means it indeed. Here is a self contained proof – which amounts more or less to prove the 1st isomorphism theorem:

Lemma: For any $g\in G$, we have $\phi^{-1}\bigl(\phi(g)\big)=g\cdot\ker\phi$.

This means any $h$ such that $\phi(h)=\phi(g)$ has the form $h=gx$, for a (unique) $x\in\ker \phi$.

Indeed, $\phi(gx)=\phi(g)\phi(x)=\phi(g)e'=\phi(g)$ ($e'$ is the unit of $G'$). Conversely, if $\phi(h)=\phi(g)$, we have $\phi(g^{-1}h)=\phi(e)=e'$, so $x=g^{-1}h\in ker \phi$ and $h=g(g^{-1}h)$.

Since $\phi$ is surjective, this proves $G$ is partitioned into the subsets $\phi^{-1}\bigl(\phi(g)\big)$ which all have cardinality $m=\lvert\ker g\rvert$. If $r$ is the number of elements in this partition, we have $$\lvert G\rvert=\underbrace{m+m+\dots+m}_{r \text{ terms}}=mr$$ by the shepherd's principle.

Bill Dubuque
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Bernard
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  • I can understand the lemma but still can't get the point how $G$ is partitioned into the subsets. And how to link this with $G'$? – Wang Kah Lun Sep 22 '16 at 13:49
  • And also $|G|\neq \sum_{g\in G}|\phi^{-1}(\phi(g))|$ right? since the mapping is not injective. – Wang Kah Lun Sep 22 '16 at 13:51
  • The subsets are the inverse images of the elements of $G'$. For your second question: you're right. This is because if $h\in \phi^{-1}\bigl(\phi(g)\big)$, then $\phi^{-1}\bigl(\phi(h)\big)=\phi^{-1}\bigl(\phi(g)\big)$ For this reason of duplicates, I speak of the set of $\phi^{-1}\bigl(\phi(g)\big)$, not of the family. – Bernard Sep 22 '16 at 14:09
  • I have updated my idea above. Can you please check whether my thought is correct? Thanks! – Wang Kah Lun Sep 22 '16 at 14:41
  • Two observations: 1) I would say ‘ the pre-images of the elements in $G'$…’, as it is not one pre-image alone which partitions $G$. 2) After hence it should be $\phi^{-1}(y)$. The end of this line supposes you have enunciated the lemma (you may call it lemma, claim or whatever you want, but you have to state it for a full justification). – Bernard Sep 22 '16 at 14:56
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As $\phi$ is an epimorphism then it is a surjective homomorphism. As such $\operatorname{im}\phi=G'$ which is the subgroup of $G'$ that, since $\phi$ is an epimorphism, is the whole of $G'$, and $\ker\phi=K$, where $K$ is the subgroup of $G$ whose elements map to $1_{G'}$, the identity in $G'$.

Since we have an epimorphism: $$|G|=|\operatorname{im}\phi||\ker\phi|=|G'||\ker\phi|$$ and $G$ is thus partitioned into $\frac{|G|}{|K|}$subsets all of order $|\ker\phi|$ which map to $G'$ under $\phi$.

Daniel Buck
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