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I am searching for an alternative definition of a group homomorphism using only their impact on subgroups. I thought of something like

Let $G$, $G'$ be groups. A map from $G$ to $G'$ is called homomorphism of groups if the image of every subgroup of $G$ is a subgroup in $G'$.

I guess this definition wouldn't work. It's maybe too weak to be a homomorphism.

Does anyone have an idea?

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    $\begingroup$ It definitely is too weak. According to your definition any bijection $\mathbb{Z}_p\to\mathbb{Z}_p$ that fixes $0$ would be a homomorphism. Why are you looking for an alternative if there is a simple classical definition? $\endgroup$
    – freakish
    Dec 3, 2018 at 9:33
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    $\begingroup$ Are you trying to do something that mimics the definition of a continuous map? $\endgroup$ Dec 3, 2018 at 9:44
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    $\begingroup$ @freakish When if first learned about normal subgroups I had absolutely no idea what this definition using conjugation really wants to tell me. It just seemed to be not very intuitive, but when I began to think of normal groups as the kernels of group homomorphism it really helped me a lot to understand what normal subgroups are all about.I hoped to find a definition which makes even clearer which structures of a group "survive" a homomorphism. $\endgroup$
    – Dominik
    Dec 3, 2018 at 9:52

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Others have pointed out the trouble with this definition. This is the way one learns anyway. Try to see why this definition and not this gives a clear and critical understanding.

Let me point out another example where your suggested alternative definition fails: Take a group which is NOT abelian, say symmeric group on $n$ letters, . $n>3$. Consider the function $x\mapsto x^{-1}$. As a function from a group to itself, it satisfies your condition, image of any subgroup is a subgroup (in fact it is the same subgroup). However it is not true in general that $(xy)^{-1}= x^{-1}y^{-1}$. So it is not a homomorphism.

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This definition is far too weak! you need compatibility with the group structure, which gets completely forgotten by your definition, for example, consider the following map: $$\mathbb{Z}/p\mathbb{Z} \to \mathbb{Z}/q\mathbb{Z}\\ x \mapsto [x]$$ where we interpret $\mathbb{Z}/p\mathbb{Z} $ as $\{0,1,...,p-1\}$. Now if $q \le p$ this is would be by your "definition" a "groupmorphism", but all the properties of the elements, like order gets messed up, and you lose all control!

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