Show that $\sigma\circ\phi : G_1 \to G_3$ is an isomorphism

Question

Let $G_1, G_2$ and $G_3$ be groups. Let $\phi: G_1 \to G_2$ and $\sigma: G_2 \to G_3$ be isomorphisms of groups. Show that $$\sigma\circ\phi: G_1 \to G_3$$ is an isomorphism.

I understand to prove the composition is a homomorphism (operation preserving) and a bijection. I need help with notation and how to show the operation preserving, onto and one-to-one.

Operation preserving: $\sigma\circ\phi(ab)=\sigma(\phi(ab))=\sigma(\phi(a)\phi(b))=(\sigma\circ\phi(a))(\sigma\circ\phi(b))$

Am I on the right track?

1

Yes you are on the right track! – R_D Oct 23 '15 at 02:41

score 3 · Accepted Answer · answered Oct 23 '15 at 02:41

Yes, now you just need that the composition of bijections is a bijection.

Perhaps by showing that $\phi^{-1}\circ \sigma^{-1}: G_3 \to G_1$ is the inverse of $\sigma \circ \phi$ (which holds for any bijections $\sigma$ and $\phi$, not necessarily between algebraic structures). Just compose this with $\sigma \circ \phi$ (on the left and on the right) to verify that you get identity maps in both cases.

score 2 · Answer 2 · answered Oct 23 '15 at 02:42

2

That is correct. All that remains is to show that the composition is a bijection.

Personally, it suffices to say "the composition of two bijections is a bijection." It is not difficult to show directly, however, either by showing subjectivity and infectivity or by showing that the inverse exists.

answered Oct 23 '15 at 02:42

pancini

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Did you mean surjectivity and injectivity instead? – Jose Arnaldo Bebita Dris Oct 23 '15 at 02:44
1

Right. Although maybe infectivity would suffice, who knows. – pancini Oct 23 '15 at 02:45
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Just realized I spelled both wrong. Damn autocorrect. – pancini Oct 23 '15 at 03:58

Show that $\sigma\circ\phi : G_1 \to G_3$ is an isomorphism

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