If $\Phi:G \rightarrow G$ is a group homomorphism, $\Phi(x)=ax^3a$ then what is $Ker \Phi, Im(\Phi)$

If $$\Phi:G \rightarrow G$$ is a group homomorphism, $$\Phi(x)=ax^3a$$ where $$a$$ is a an element in $$G$$, then what is $$Ker \Phi, Im(\Phi)$$

The question asked to show that

$$Ker \Phi = \{x\in G | x^2 = e\}$$ $$Im \Phi = \{x^2 | x\in G \}$$

I've tried calculating $$a^2$$ first and found it to be $$a^2=e$$. So the kernel is

$$Ker \Phi = \{x\in G | ax^3 a = e\}$$ $$Ker \Phi = \{x\in G | x^3 = e\}$$

However, I don't see how I can reduce this to proving $$x^2=e$$ ?

For the image, I have

$$Im \Phi = \{y | \exists x\in G , \quad y=ax^3a \}$$ $$Im \Phi = \{y | \exists x\in G , \quad aya=x^3 \}$$ $$Im \Phi = \{y | \exists x\in G , \quad ay^3a=x^9 \}$$ $$Im \Phi = \{y | \exists x\in G , \quad \Phi(y)=x^9 \}$$

But I don't see how to get it down to the form too. Any help?

• I think you may have dropped an inverse in the definition of $\Phi$: I suspect it should be $\Phi(x) = a x^3 a^{-1}$. As the answer below points out, as it stands $\Phi$ is not a homomorphism since for instance $\Phi(e) = a e^3 a = a^2 \neq e$. Dec 14 '18 at 5:07
• @André3000 I think the title intends to say that $\Phi$ is a homomorphism by hypotehsis. But even so (if my fixed argument is now okay) the conclusion does not generally follow. Dec 14 '18 at 5:41

Suppose that $$a = 1$$. Then, $$\Phi$$ is a morphism if $$(xy)^3 = x^3y^3$$ which occurs in particular if $$G$$ is abelian. In this case, $$\Phi(x) = x^3$$ and so

$$\ker \Phi = \{x \in G : x^3 = 1\} \text{ and } \operatorname{im} \Phi = \{x^3 : x \in G\}.$$

Consider then $$\mathbb{Q}^\times = \mathbb{Q} \setminus \{0\}$$ with $$q \cdot r := qr$$. Here, the element $$4 = 2^2$$ is not in $$\operatorname{im} \Phi$$, since $$4$$ is not $$q^3$$ for some fraction $$q$$.

If on the other hand we take $$\mathbb{R}^\times = \mathbb{R} \setminus \{0\}$$ and $$x \cdot y := xy$$, then $$\ker \Phi = \{x \neq 0 : x^3 = 1\} = \{1\} \subset \mathbb{R}$$

which differs from $$\{x : x^2 = 1\} = \{1,-1\}$$.

Therefore both claims are false for arbitrary $$a$$ and $$G$$.

• A remark: if $R$ is a ring, $R^\times$ typically denotes the group of units of $R,$ not the nonzero elements (which isn't even a monoid under multiplication unless $R$ is an integral domain). In particular, this does not agree with your notation. More importantly, $\Bbb Z\setminus\{0\}$ isn't a group under multiplication, though. Dec 14 '18 at 5:12
• @Stahl Well, this is embarassing... I surely need some more caffeine! Thanks for pointing this out! Dec 14 '18 at 5:34
• @Stahl I have edited the post to mean what I intended to say originally. I wouldn't mind a sanity check, though, in case my argument is still wrong. Dec 14 '18 at 5:39
• Looks good to me now! Dec 14 '18 at 5:43