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?

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

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, $g\in Y$.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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