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I have the following problem

Let $G$ and $G_1$ be finite groups such that $\gcd(|G|,|G_1|)=1$, and let $\phi:G\to G_1$ be a group homomorphism. Prove that $ker \phi=G$.

We can assume $\ker\phi$ is a subgroup of $G$.

My attempted proof is as follows:

enter image description here

It would be highly appreciated if someone could help to prove this as I am preparing for a group theory test. Thanks :)

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  • $\begingroup$ I forgot to mention that we cannot use the Isomorphismm theorems $\endgroup$
    – Amine
    Apr 19 '17 at 20:04
  • $\begingroup$ Hi, thanks for your help everyone, all the solutions were great and useful. Unfortunately I can only tick one. :( $\endgroup$
    – Amine
    Apr 19 '17 at 20:41
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Here is a proof using only basic facts.

Take any $x\in G$, set $y=\phi(x)$. We have $x^{|G|} = e_G$, and additionally $y^{|G_1|} = e_{G_1}$, by a simple form of Lagrange. Applying $\phi$ to the first equation, we have $y^{|G|} = y^{|G_1|} = e_{G_1}$.

Since $|G|$ and $|G_1|$ are relatively prime, there are integers $a,b$ such that $a|G| + b|G_1| = 1$. Then $y = y^{a|G| + b|G_1|} = (y^{|G|})^a (y^{|G_1|})^b = e_{G_1}^a e_{G_1}^b = e_{G_1}$, i.e. $x\in\ker\phi$.

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  • $\begingroup$ Hi, your solution makes sense. Just to clarify on notation; $x^(|G|)$ $=|G|x(|G|)^(-1)$ ? $\endgroup$
    – Amine
    Apr 19 '17 at 20:28
  • $\begingroup$ @Amine No, I mean $x$ raised to the power of the cardinality of $G$. I'm not sure what the meaning of the expression you wrote is. $\endgroup$
    – Slade
    Apr 19 '17 at 20:36
  • $\begingroup$ Ok cool, thanks for your help $\endgroup$
    – Amine
    Apr 19 '17 at 20:38
  • $\begingroup$ Why is $y^{|G|}=e_{G_1}$? In the last line $\endgroup$
    – Amine
    Apr 20 '17 at 5:38
  • $\begingroup$ This was proved in the second paragraph. $\endgroup$
    – Slade
    Apr 20 '17 at 21:06
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First a comment on your proof. $\phi(H)\subset H$ is not true. We have $\phi(H)\le G_1$, that is the image of $H$ is a subgroup of $G_1$, and unless $G\cap G_1\supset H$, we can't have $H\le G_1$. (But since the orders of $G$ and $G_1$ are coprime this cannot occur).

I will give a simple proof of this using the first isomorphism theorem. Notice that $\phi(G)\cong G/\ker\phi$. Thus, $|\phi(G)||\ker\phi|=|G|$. But $\phi(G)\le G_1$, so the order of $\phi(G)$ divides the order of $G_1$. Thus, $|\phi(G)|$ divides both $|G_1|$ and $|G|$. Since these are coprime, it follows that $|\phi(G)|=1$, and thus $|\ker(\phi)|=|G|$, so $\ker\phi=G$.

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  • $\begingroup$ Hi, Thanks for the response, I forgot to mention that we cannot use the Isomorphismm theorems. Actually this is why I am stuck. Is it possible to prove it without the use of the theorem? $\endgroup$
    – Amine
    Apr 19 '17 at 20:02
  • $\begingroup$ Are you allowed to use Lagrange's Theorem? $\endgroup$ Apr 19 '17 at 20:07
  • $\begingroup$ Yes I am allowed to $\endgroup$
    – Amine
    Apr 19 '17 at 20:16
  • $\begingroup$ It looks like you already have a response using the tools you are allowed, so I will leave my answer as is. $\endgroup$ Apr 19 '17 at 20:23
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You should know (or can verify) that if $\phi$ is a group homomorphism, then $|\phi(x)|$ divides $|x|$. Hence, for all $x$, $\phi(x)$ must be chosen to fit this qualification. But then $|\phi(x)|$ divides both $|G|$ and $|G_1|$, so $|\phi(x)|=1$, so $\phi(x)$ is the group identity in $G_1$.

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  • $\begingroup$ Is $x\in \phi(x)$? Can you give me a tip on how I can show $|\phi(x)| divides |x| and why is it useful in this case? $\endgroup$
    – Amine
    Apr 19 '17 at 20:22
  • $\begingroup$ $x$ and $\phi(x)$ are elements of $G$ and $G_1$ respectively, so one can't "contain" the other. If $|x|=n$, then $x^n=e_G$. Furthermore, $\phi(x^n)=\phi(x)^n=e_{G_1}$ since a homomorphism must preserve the identity. Then, by definition of the order, the order of $\phi(x)$ must divide $n=|x|$. If $|\phi(x)|$ divides $|x|$, then by Lagrange's theorem, since $|x|$ divides $|G|$, it follows that $|\phi(x)|$ divides $|G|$. Lagrange's theorem also tells you that $|\phi(x)|$ divides $|G_1|$. $\endgroup$
    – Kevin Long
    Apr 19 '17 at 20:29

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