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The problem is :

Let $H$ be a normal subgroup of $G$ of order $6$.If $f : G \longmapsto G_1$ be an epimorphism of groups such that $H \subseteq ker f$, then show that $G_1$ is also a homomorphic image of $G/H$.

I don't find any route of solving this problem.I don't even understand the significance of the fact that $|H| = 6$.Please help me by giving me a hint at least.Then I will retry it.Thank you in advance.

  • The order 6 is indeed irrelevant. Ignore it – Hagen von Eitzen Jul 05 '16 at 06:28
  • You want to construct a homomorphism $\phi:G/H\to G_1$. You already know that $f$ sends elements of $g$ to $G_1$. So the natural thing to try is to send the coset $gH\in G/H$ to $f(g)$... – symplectomorphic Jul 05 '16 at 06:33
  • Oh! Nice idea indeed.But the given condition is useless which confused me.Because I entirely tried to focus myself on the fact that $|H| = 6$ which misguided me. –  Jul 05 '16 at 06:42

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Hint. Is $gH\mapsto f(g)$ well-defined?

Hagen von Eitzen
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    Let $g_1H = g_2H$,for some $g_1,g_2 \in H$.Then $g_1^{-1}g_2 \in H$.Since, it is given that $H \subseteq ker f$.So,$g_1^{-1}g_2 \in ker f$.So $f(g_1^{-1}g_2) = e$, where e is the identity element in $G_1$.Since $f$ is a homomorphism it follows that $f(g_1) = f(g_2)$ i.e. $\varpi (g_1H) = \varpi (g_2H)$.Hence $\varpi$ becomes well defined. –  Jul 05 '16 at 07:45
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    Sorry $g_1,g_2 \in G$.Excuse me please. –  Jul 05 '16 at 07:50