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The problem is:

Let G and G' be groups, and let H and H' be normal subgroups of G and G', respectively. Let $\phi$ be a homomorphism of G into G'. Show that $\phi$ induces a natural homomorphism

$$\phi_*: (G/H) \rightarrow (G'/H)\ \text{if}\ \phi[H] \subseteq H'.$$ (This fact is constantly used in algebraic topology.)

Attempt at a solution:

$\phi: G \rightarrow G'$

$\phi_*(gH) = \phi(g)H'$ where $gH$ is the class of $g$ in $\frac{G}{H}$ and $\phi(g)H'$ is the class of $\phi(g)$ in $\frac{G'}{H'}$.

$$\phi_*((aH)(bH)) = \phi_*((ab)H)$$

$$ = \phi(ab)H'$$

$$= \phi(a)\phi(b)H'$$

$$= \phi(a)H'\phi(b)H'$$


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2 Answers 2

up vote 1 down vote accepted

Ok, but you have to verify that $\phi_*$ is well-defined. In order not to repeat the same computations all the time, remember the homomorphism theorem:

If $H$ is a normal subgroup of $G$, there is a surjective homomorphism $p : G \to G/H$ with kernel $H$. If $K$ is a group, then every group homomorphism $f : G \to K$ which vanishes on $H$ has a unique extension to a homomorphism $\overline{f} : G/H \to K$, i.e. $\overline{f} \circ p = f$.

Here, you can use the composition $f : G \stackrel{\phi}{\to} G' \to G'/H'$. The assumption $\phi(H) \subseteq H'$ means exactly that $f$ vanishes on $H$. Thus, it extends to a homomorphism $G/H \to G'/H'$. No computation is needed at all. In general, it is good to forget about cosets at all and just remember the universal property of $p : G \to G/H$ above.

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Looks fine, but you missed to check well-definedness ... that is: If $g, g' \in G$ have the same coset modulo $H$, $gH = g'H$, then the image of this coset has to be unabiguously defined, so we must have $\phi_*(gH) = \phi_*(g'H)$, meaning $\phi(g)H' = \phi(g')H'$ for $g,g' \in G$ with $gH = g'H$.

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