# All groups $(G, \cdot)$ such that $\forall g_1, g_2 \in G, g_1,g_2 \in G \iff g_1 \cdot g_2 \in G$?

I want to find a group $$(G, \cdot)$$ (group $$G$$ with a binary operation $$\cdot$$) such that for all elements $$g_1, g_2 \in G$$, the following holds:

$$g_1 \cdot g_2 \in G \iff g_1, g_2 \in G \tag{1} \label{1}$$

The converse is always true because it is one of the definitions of being a group. That is,

Closure: If $$g_1, g_2 \in G$$, then $$g_1 \cdot g_2 \in G$$, for all $$g_1, g_2 \in G$$.

It seems impossible because there is a counterexample in this post I wrote.

Is this possible? That is, is there a group $$(G, \cdot)$$ for $$\eqref{1}$$ to be true?

I tried using conjugacy classes, but conjugacy classes are just sets. Any suggestions?

Thanks.

• Your formula (1) is obviously incomplete. Jul 30, 2022 at 19:14
• That's true, let me edit it. My bad. I forgot to write the whole equation. Jul 30, 2022 at 19:15
• No, there isn't, because that post lacks the quantifier "for all $g_1,g_2$ IN $G$." In this post you start by telling us we cannot go outside of $G$. Jul 30, 2022 at 19:21
• Your old post was like saying "Is it true a person is alive if and only if they are living in the United States?" and the answer was of course "no". Now you are asking "Is it true that every person in the United states is alive if and only if they are living in the United States?" and that is a different question with an affirmative answer, because you start by restricting yourself only to people in the United States, as opposed to the original question where you did mot. Jul 30, 2022 at 19:27
• Several of us explained that in comments there... Jul 30, 2022 at 19:31

You are certainly right that $$g_1,g_2 \in G\Rightarrow g_1\cdot g_2 \in G$$ is apart of the axioms for a group and so is correct. The issue lies with the opposite statement $$g_1\cdot g_2 \in G\Rightarrow g_1,g_2 \in G$$ which, without further context, is nonsensical. To try and explain this, ask yourself: where are $${g_1,g_2}$$ being taken from? What group or set do they belong to? If your answer to that is "they come from $$G$$" then clearly $${g_1,g_2 \in G}$$, but this is a rather trivial statement at this point (you are just saying $${g_1,g_2 \in G\Rightarrow g_1,g_2 \in G}$$). So in order for us to have a non-trivial statement, $${g_1,g_2}$$ must be coming from somewhere outside of $$G$$, but you have not been specific in describing exactly where they come from and how $$G$$ comes into this at all. Hopefully that helps, if not let me know and I'll try to explain further

EDIT: my mistake, I didn't see in the post you did indeed specify where $${g_1,g_2}$$ were to come from, which was indeed $$G$$. As I say though, this makes the full statement $$\forall\ g_1,g_2 \in G: g_1\cdot g_2 \in G\Rightarrow g_1,g_2 \in G$$ trivially true since you are now specifying that $${g_1,g_2}$$ should come from $$G$$.

Riemann'sPointNose has given the answer to your question as written (+1), which is that this is trivially true! Namely, the following statement is trivially true:

Let $$(G,\cdot)$$ be a group. Then for all $$g_1,g_2\in G$$ such that $$g_1g_2\in G$$ we have $$g_1,g_2\in G$$.

Indeed, you start by assuming that $$g_1,g_2$$ lie in $$G$$, and then wish to conclude that $$g_1,g_2$$ lie in $$G$$. Do you see why this is a vacuously true statement? The fact that $$g_1g_2\in G$$ has nothing to do with it, it just follows from the assumption that $$g_1,g_2\in G$$ to begin with.

However, there's a different way of interpreting your question, which perhaps is more in the spirit of what you were looking for:

For which groups $$(G,\cdot)$$ is the following true? Whenever $$\alpha:G\to G'$$ is an injective homomorphism (ie whenever $$G$$ is "morally" a subgroup of $$G'$$), we have $$ab\in \alpha(G)\implies a,b\in \alpha(G)$$ for all $$a,b\in G'$$.

It turns out that there is no group with this property. In fact, for any homomorphism $$\alpha:G\to G'$$ of groups, provided that $$G'\neq\alpha(G)$$ we can always find $$a,b\in G'\setminus\alpha(G)$$ such that $$ab\in\alpha(G)$$! Indeed, just pick any $$t\in G'\setminus\alpha(G)$$, and let $$a=t$$ and $$b=t^{-1}$$. So to show that a group $$G$$ does not have this property we just need to find any group $$G'$$ with an injective homomorphism $$\alpha:G\to G'$$ such that $$G'\neq\alpha(G)$$; for example, we can take $$G'=G\times\mathbb{Z}$$ and $$\alpha(g)=(g,0)$$.

I am not sure whether this is a satisfactory answer for you, but thought I'd point it out, as it's in some sense the natural follow-up to the question you asked.

• I really like this follow up question actually! I did think that maybe the question could be rephrased instead to talk about embedding $G$ in some other group in this sorta way, but didn't manage to think of any concrete statement with solution like this. Nice! (+1) Jul 30, 2022 at 22:20