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?

  • $\begingroup$ 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 $\Phi(e) = a e^3 a = a^2 \neq e$. $\endgroup$ Dec 14 '18 at 5:07
  • 2
    $\begingroup$ @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. $\endgroup$
    – qualcuno
    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 $G$.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Stahl
    Dec 14 '18 at 5:12
  • $\begingroup$ @Stahl Well, this is embarassing... I surely need some more caffeine! Thanks for pointing this out! $\endgroup$
    – qualcuno
    Dec 14 '18 at 5:34
  • $\begingroup$ @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. $\endgroup$
    – qualcuno
    Dec 14 '18 at 5:39
  • 1
    $\begingroup$ Looks good to me now! $\endgroup$
    – Stahl
    Dec 14 '18 at 5:43

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