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Given a group $G$ , and two subgroups $ G_1, G_2 $ such that $ G_1 \subseteq G_2 $ , is it true that $ G/G_2 \subseteq G/G_1 $ ?

Thanks in advance !

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If you know what $G/G_1$ and $G/G_2$ are, then you can use the subgroup test to answer this. –  Arkamis Jul 17 '12 at 16:51
    
But is it always true? –  joshua Jul 17 '12 at 17:04
    
It would be if you can show it generally. Is this a homework problem, by chance? –  Arkamis Jul 17 '12 at 17:05
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2 Answers 2

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What you probably had in mind is the following: $$G/G_2\cong\left(G/G_1\right)/\left(G_2/G_1\right)$$ by the 2nd or 3d. isomorphism theorem, so what you asked can't generally be as $\,G/G_2\,$ is a quotient of $\,G/G_1\,$ , i.e.: there exists a surjective group homomorphism $$G/G_1\longrightarrow G/G_2\,\,,\,\,xG_1\to xG_2$$

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Grear! Thanks!! –  joshua Jul 17 '12 at 18:45
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The quotients may not make sense, if $G_1$ and $G_2$ are not normal.

However, it is true that every left-$G_2$ coset is completely contained in a $G_1$ coset. That is, the equivalence relation induced by the larger subgroup is "coarser" than that of the larger group. In other words, $x{}_{G_1}\equiv y$ implies $x{}_{G_2}\equiv y$, but not necessarily conversely.

To see this, simply remember that $x{}_{G_i}\equiv y$ if and only if $y^{-1}x\in G_i$. If $x{}_{G_1}\equiv y$, then $y^{-1}x\in G_1\subseteq G_2$, so $x{}_{G_2}\equiv y$. But if $G_2$ properly contains $G_1$, then let $x\in G_2-G_1$; then $x{}_{G_2}\equiv e$, but $x{}_{G_1}\notequiv e$.

Similar results hold for congruent-on-the-right (equivalently, right cosets).

If both $G_1$ and $G_2$ are normal, so that the set of equivalence classes can be given a group structure, then the coarser equivalence relation induces a smaller group, so that $G/G_2$ is a quotient of $G/G_1$.

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