# Showing that if a subgroup is normal, it's homomorphic image is normal

$\alpha : G \to H$ is a surjective homomorphism. And $U \subset G$ is a subgroup of $G$. Verfiy the claim -

The image of $U$, ie $\alpha(U)$, is a subgroup of $H$, and if $U$ is normal in $G$, then $\alpha(U)$ is normal in $H$.

Answer:

Firstly, do I have to show $\alpha(U)$, is a subgroup of $H$ or is that statement just a statement of fact as part of the question?

Anyway here is what I have done..taking it as a given that $\alpha(U)$, is a subgroup of $H$ -

As $U$ is normal we have

$U = gUg^{-1}$

$\alpha(U) = \alpha(gUg^1) =$ {applying homomorphic mapping into H} = $\alpha(g)\alpha(U)\alpha(g^{-1})$

Is that correct? I have a feeling I should take the $-1$ exponent outside the bracket as an extra final step or is that superfluous?

-

1
The hypotheses should include the statement that $U$ is a subgroup of $G$. You must verify (1) that $\alpha[U]$ is a subgroup of $H$, and (2) that if $U$ is normal in $G$, than $\alpha[U]$ is normal in $H$. Yes, you need to finish the proof of (2) by pulling the inverse outside, so that the last factor is $(\alpha(g))^{-1}$. – Brian M. Scott Oct 23 '12 at 0:04
1. Yes, if you are uncertain, try to prove it. 2. Yes, final step, inverse out. – Berci Oct 23 '12 at 0:05
@BrianM.Scott: Yes I have edited in that it $U$ is a subgroup of $G$. – Jim_CS Oct 23 '12 at 0:06

## 2 Answers

To show that $\alpha(U)$ is a normal subgroup, you need to prove that $\alpha(U) = x \alpha(U) x^{-1}$ for all $x \in H$. But any $x \in H$ can be written in the form $\alpha(g)$ for $g \in G$ since $\alpha$ is surjective. Thus, you need only prove that $\alpha(U) = \alpha(g) \alpha(U) \alpha(g)^{-1}$ for all $g \in G$. Note the $-1$ exponent is "on the outside", so you really do need to take that last step as you suspected.

And yes, as written it is meant that you should prove that $\alpha(U)$ is a subgroup. The proof requires you to carefully work through the definition/criterion for being a subgroup, but nothing beyond that.

Also, +1 for showing your reasoning and clearly indicating what you are unsure of.

-

I think you must have meant $\,\alpha(U)=x\alpha(U)x^{-1}\,$ for all $\,x\in H\,$ ... – DonAntonio Oct 23 '12 at 2:53
Yes, thanks. :)  I've corrected the mistake. – Michael Joyce Oct 23 '12 at 3:35

You have to prove that $\alpha(U)$ is normal subgroup in $H$, so for all $x\in H$, one has $x\cdot \alpha(U)\cdot x^{-1} \subseteq \alpha(U)$.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 47, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9657390713691711, "perplexity_flag": "head"}