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Let's say we have $K \subset H \subset G$, for context, let us identify these objects are groups. We shall also assume, in this case, that $K,H$ are normal in $G$. Then does it make sense to talk about the quotient $K/H$? Similarly, can one talk about $H/K$ if $K$ is only normal in $G$? Because the quotient map $\phi : H \to H/K$ dictates that $\ker \phi = K$, so appears so.

Yet $\pi : H \to K/H$ seems to suggest $\ker \pi = H$ since identity is the same for all 3 groups.

jacob smith
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  • No it doesn't make sense to talk about $K/H$. Do you understand what a quotient group is? – anon Sep 02 '16 at 22:23
  • It's just a group partitioned by the thing you are modding it out. Roughly speaking. But I know where the wrong is from the answer below. – jacob smith Sep 02 '16 at 22:27

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If $K\subset H \subset G$, then no assumptions about normality will help the expression $K/H$ make sense; it just doesn't mean anything.

However, even if $K$ is not normal in $G$, it may still be normal in $H$, in which case you can still form the quotient group $H/K$.

You seem to be a little confused about how quotient homomorphisms work. Given a group $A$ and a normal subgroup $B\trianglelefteq A$, there is a quotient homomorphism $q:A\to A/B$, defined by $q(a)=aB$, and $\ker(q)=B$. Neither of the maps $\phi:K\to H/K$ or $\pi:H\to K$ fit this description, or even really make sense; what is their explicit definition supposed to be.

Zev Chonoles
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  • But I am asking if it makes sense to talk about $K/H$, not $H/K$ – jacob smith Sep 02 '16 at 22:10
  • @jacobsmith: Sorry, I've added to my answer addressing that. – Zev Chonoles Sep 02 '16 at 22:14
  • Right, I forgot the quotient map demands $B \subset A$. Such a containment is impossible by my construction. And an obvious corollary from this states $B \subset A$ for any quotient structure $A/B$. – jacob smith Sep 02 '16 at 22:16
  • This is also another question simple, but not totally unrelated. Is the preimage of a surjective map also surjective? In context, I am referring to the quotient map. I would assume so just from the definition. – jacob smith Sep 02 '16 at 22:18
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    @jacobsmith: Well, given a function $f: X\to Y$, you can talk about the preimage of a subset $Z\subset Y$ under the function $f$, making the subset $f^{-1}(Z)\subset X$, but it doesn't really make sense to talk about the preimage of a map. – Zev Chonoles Sep 02 '16 at 22:20
  • I thought maps and functions are the same thing? – jacob smith Sep 02 '16 at 22:21
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    @jacob Yes, maps and functions are the same thing. What's your point? A preimage is a set, it doesn't make sense to call a set surjective. – anon Sep 02 '16 at 22:22
  • @jacobsmith: They are, my point is that the concept of "preimage" requires two inputs, a function $f: X\to Y$ and a subset of $Y$, and it produces an output that is a subset of $X$. Whereas you asked about "preimage of a surjective map" without specifying a subset, and asked whether the result "is surjective", which is not a property of sets. – Zev Chonoles Sep 02 '16 at 22:24
  • Well I pose this way. If $f: X \to Y$ is a function and onto, then is $f^{-1}: Y \to X$ onto? So if $U \subset X$ and $V \subset Y$, then if the restriction map $f : U \to Y$ is surjective such that $f(U) = V$, then is the inverse map $f^{-1}(V) = U$? – jacob smith Sep 02 '16 at 22:30
  • @jacobsmith: Not every surjective function $f$ has an inverse function $f^{-1}$. The symbol $f^{-1}$ is used in the context of preimages and also inverse functions, and there is a certainly a relationship between those two uses, but $f:X\to Y$ being a surjective function is not enough to talk about a function $f^{-1}:Y\to X$; the function $f$ has to be bijective. – Zev Chonoles Sep 02 '16 at 22:33
  • However, if maybe I understand your intention, it is true that if $f:X\to Y$ is surjective, and $U$ is any subset of $X$, and $V=f(U)$ is the image of $U$ under the function $f$, then the restricted function $f:U\to V$ is also surjective. – Zev Chonoles Sep 02 '16 at 22:34
  • @ZevChonoles, right. Maybe if I ask you if $f^{-1}(V) = U$ is true only given that $f$ is surjective? – jacob smith Sep 02 '16 at 22:34
  • @jacobsmith: Okay, I think I understand your question. For any function $f:X\to Y$, the statement $U\subseteq f^{-1}(f(U))$ is always true for any subset $U$ of $X$, and we have $U=f^{-1}(f(U))$ for any subset $U$ of $X$ if and only if $f$ is injective. This math.SE thread may help you out: http://math.stackexchange.com/q/359693/264 – Zev Chonoles Sep 02 '16 at 22:38