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Is my proof ok?

Let $f:G\to G^{\prime}$ be a group homomorphism and let $H\leq G$. $\text{Im}(H) = \{f(x):x\in H\}$. To show that $\text{Im}(H)$ is a group, it suffices to show that $f(x)f(y)^{-1}\in \text{Im}(H)$. $$f(x)f(y)^{-1}=f(xy^{-1})\in \text{Im}(H)\text{ because } xy^{-1}\in H$$

$\text{Ker}(f)=\{x:f(x)=e^{\prime}\}$ ($e^{\prime}$ is the identity in $G^{\prime}$). $$f(xy^{-1})=f(x)f(y)^{-1}=e^{\prime}(e^{\prime})^{-1}=e^{\prime}\in \text{Ker}(f)$$.

MyMathYourMath
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mrk
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3 Answers3

9

It's ok but don't forget to show that $\mathrm{Im}(H)\ne \emptyset$ since $f(e)=e'\in\mathrm{Im}(H)$ and the same thing for $\ker f$.

5

Yes, your proofs are correct. To be perfectly rigorous you also need to say that these sets are nonempty (which is obvious, but still).

Najib Idrissi
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Reference: http://www.proofwiki.org/wiki/Group_Homomorphism_Preserves_Subgroups

Duplicates: If $K$ is a subgroup of $G$, then $\phi(K) = \{ \phi(k) | k \in K \}$ is a subgroup of $\bar{G}.$ Vice-versa? Why is the image of a homomorphism a subgroup of the codomain? Showing the image of a subgroup is a subgroup.

== Theorem ==

Let $\left({G_1, \circ}\right)$ and $\left({G_2, *}\right)$ be [[Definition:Group|groups]].

Let $\phi: \left({G_1, \circ}\right) \to \left({G_2, *}\right)$ be a [[Definition:Group Homomorphism|group homomorphism]].

Then: : $H \le G_1 \implies \phi \left({H}\right) \le G_2$ where $\le$ denotes [[Definition:Subgroup|subgroup]]. That is, [[Definition:Group Homomorphism|group homomorphism]] preserves [[Definition:Subgroup|subgroups]].

Proof: Let $H \le G_1$. First note that from [[Null Relation is Mapping iff Domain is Empty Set]]: : $H \ne \varnothing \implies \phi \left({H}\right) \ne \varnothing$

and so $\phi \left({H}\right)$ is not [[Definition:Empty Set|empty]].

Next, let $x, y \in \phi \left({H}\right) \iff \exists h_1, h_2 \in H: x = \phi \left({h_1}\right), y = \phi \left({h_2}\right)$

We want to use the the [[One-Step Subgroup Test]]. Hence need $x*y^{-1} \in \phi(H) :$

$x*y^{-1} = \phi(h_1)*[\phi(h_2)]^{-1} = \phi(h_1)*\phi(h_2^{-1}) = \phi(h_1\circ h_2^{-1}).$

Since $H$ is a [[Definition:Subgroup|subgroup]], $h_1\circ h_2^{-1} \in H$. Hence $\phi(h_1\circ h_2^{-1}) \in \phi \left({H}\right) \le G_2$.

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