0
$\begingroup$

Question:

Suppose that $\alpha: G \to H$ is a surjective homomorphism of groups. Let $U$ be a subgroup of $H$ and prove the following:

The pre-image of $U$ under $\alpha$, .ie. $\{g \in G | g^{\alpha} \in U\}$, is a subgroup of $G$ containing $\ker(\alpha)$.

My answer:

Define $Y = \{g \in G | g^{\alpha} \in U\}$.

$1$. First showing $Y$ is a subgroup using the one-step subgroup test.

Let $g_1, g_2 \in Y$.

$(g_1g_2)^\alpha = g_1^\alpha g_2^\alpha \in U$

So, $g_2^\alpha \in U \implies (g_2^\alpha)^{-1} \in U \implies (g_2^{-1})^\alpha \in U \implies g_2^{-1} \in Y$

So we have $g_1g_2^{-1} \in Y$. And $Y$ is non-empty as $I_H \in U \implies I_G \in Y$ as homomorphisms preserve the identity element. Therefore $Y$ is a subgroup.

$2$. Now showing $\ker(\alpha) $ is contained in $Y$.

We have $I_H \in U$.

$\implies$ there exists $g_1, g_2,...,g_n \in Y$ such that $g_i^\alpha = I_H$, $i = 1,...,n$

So the set $\{g_1, g_2,...,g_n\} = \ker(\alpha)$ and hence $\ker(\alpha)$ is contained in $Y$.

How is that answer?

$\endgroup$
  • $\begingroup$ Note that one doesn't actually need the surjectivity hypothesis. $\endgroup$ – anon May 21 '15 at 18:00
1
$\begingroup$

It looks good for the most part, though I think your justification for $\ker(\alpha)\subseteq Y$ is a bit shaky (the kernel need not be finite). You'll want to take an arbitrary member of $\ker(\alpha)$, say $g$, and show that $g\in Y$. Indeed, $g^\alpha=I_H\in U$, so by definition, $g\in Y$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, I was wondering about that myself but I didn't know how to write down that it can be finite or infinite. If I said $g_1, g_2,... \in Y$ then I am saying that the kernel is infinite and leaving out that it could be finite. Is there a way of notating it that says it could be finite or infinite? $\endgroup$ – sonicboom Nov 14 '12 at 21:16
  • 1
    $\begingroup$ Note that the approach I took completely avoids reference to the cardinality of $\ker(\alpha)$. We took $g\in\ker(\alpha)$, and showed that $g\in Y$. Since $g$ was arbitrary, then for all $g\in\ker(\alpha)$ we have $g\in Y$--that is, $\ker(\alpha)\subseteq Y$, as desired. This is a fairly standard approach when we need to show that $A\subseteq B$ for some $A,B$: take $a\in A$, and show that $a\in B$. $\endgroup$ – Cameron Buie Nov 14 '12 at 21:23
  • $\begingroup$ It seemed to me though if I said - there exists $g \in Y$ such that $g^\alpha = I_H$ that I am saying that there is only one $g$ that maps to the identity of $H$. $\endgroup$ – sonicboom Nov 14 '12 at 21:29
  • 1
    $\begingroup$ I can see why you'd interpret it that way. What that says is that there is at least one $g$ that maps to the identity of $H$. That isn't what we are trying to show, though. We're trying to show that every $g$ that maps to $I_H$ is in $Y$. $\endgroup$ – Cameron Buie Nov 14 '12 at 21:33
  • $\begingroup$ Ok I get it now, cheers. I am putting down - Let $g \in ker(\alpha)$. As $g^\alpha = I_H \in U \implies g^\alpha \in Y$ and hence $ker(\alpha) \in Y$. $\endgroup$ – sonicboom Nov 14 '12 at 21:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.