4
$\begingroup$

This is a question I attempted from a text book. And I would like to know how far I am correct in my arguments and thought process.

Let $H$ be a subgroup of a group $G$, and let $\phi: G \to H $ be a homomorphism whose restriction to $H$ is the identity map: $\phi(h) = h$ , if $h \in H$. Let $N = ker(\phi)$

(a) Prove that if $G$ is abelian then it is isomorphic to the product group $H \times N$. (b) Find a bijective map $G\to H \times N$ without the assumption that $G$ is abelian, but show by an example that $G$ need not be isomorphic to the product group.

My attempt: (a) Aim is show a bijection and homomorphism between $G$ and $H \times N$. Denote the isomorphism as $\psi : H \times N \to G$

First looking at the homomorhism $\phi$, I see that it's surgective homomorphism since each element in the codomain $H$ is mapped at least to $H$ subgroup in $G$.

We have $\phi: G \to H $, i.e., the the whole group has $H$ as it's image which is also the image of $H$ subgroup in $G$. $\phi(H)=H$ and $\phi^{-1}(\phi(H))=G=NH$(also $=HN$ since $N$ is a normal subgroup). Thus $G = HN$ which assures the surgectivity of $\psi$

Next observation is that $H \cap N$ is $\{1\}$ since $\phi(h) = h$ , if $h \in H$. Consider elements $(h_1,n_1), (h_2,n_2)$ in $H \times N$ which have same image $h_1n_1=h_2n_2$ in $HN$. i.e., $h_1h_2^{-1}=n_1^{-1}n_2$. So, $h_1h_2^{-1}=n_1^{-1}n_2 = 1$. Hence $h_1=h_2$ and $n_1=n_2$. Injectivy of $\psi$ is shown.

Now consider the composition of two elements in $H \times N$ $(h_1,n_1)$ and $(h_2,n_2)$, $(h_1h_2,n_1n_2)$. It's image in $G=HN$ would be $h_1h_2n_1n_2$ while the composition of individual products in $G$ would be $h_1n_1h_2n_2$. Since $G$ is abelian we say these are equal. Hence we have a homomorphism along with bijection making it isomorhism for $\psi: H \times N \to G$ .

(b) If $G$ is not abelian, we don't have homomorphism, but we still have bijection $\psi: H \times N \to G$ defined by $\psi(h,n)=hn$

Please let me know if there are any incorrect/loose arguments. Thanks

$\endgroup$
3
$\begingroup$

I don't see where you have defined $\psi$. I presume your intended map is:

$\psi(h,n) = hn$. I don't see how you establish this is surjective.

What I would do is this, define:

$\psi: G \to H \times N$ by $\psi(g) = (\phi(g), \phi(g)^{-1}g)$

It may not be clear the second coordinate is in $N$, so let's demonstrate:

$\phi(\phi(g)^{-1}g) = \phi(\phi(g^{-1}))\phi(g) = \phi(g^{-1})\phi(g) = \phi(e) = e$

(because $\phi(g^{-1}) \in \text{im }\phi = H$, and $\phi$ is the identity on $H$), so $\phi(g)^{-1}g \in \text{ker }\phi = N$.

Now we can establish bijectivity of $\psi$. Suppose $\psi(g) = \psi(g')$.

Then $\phi(g) = \phi(g')$, and $\phi(g)^{-1}g = \phi(g')^{-1}g'$.

So the second equation becomes: $\phi(g)^{-1}g = \phi(g)^{-1}g' \implies g = g'$.

On the other hand, it is clear that $\psi(hn) = (\phi(hn),\phi(hn)^{-1}(hn))$

$= (\phi(h)\phi(n),\phi(n^{-1})\phi(h^{-1})hn) = (h,\phi(n)^{-1}h^{-1}hn) = (h,n)$, so $\psi$ is surjective.

Now the question becomes: is $\psi$ a homomorphism?

$\psi(gg') = (\phi(gg'), \phi(gg')^{-1}gg') = (\phi(g)\phi(g'),\phi(g')^{-1}\phi(g)^{-1}gg')$.

IF $G$ is Abelian, then $\phi(g')^{-1}\phi(g)^{-1}gg' = \phi(g)^{-1}g\phi(g')^{-1}g$, and we continue:

$= (\phi(g)\phi(g'), \phi(g)^{-1}g\phi(g')^{-1}g) = (\phi(g),\phi(g)^{-1}g)(\phi(g'),\phi(g')^{-1}g'))$

$=\psi(g)\psi(g')$.

To show (b) you need a COUNTER-EXAMPLE.

Let $G = D_3 = \{e,a,a^2,b,ba,ba^2\}$, the dihedral group of order 6.

Let $H = \{e,b\}$ and $N = \{e,a,a^2\}$. Define $\phi: G \to H$ by:

$\phi(a^kb^j) = b^j$ for $k = 0,1,2, j = 0,1$. I leave it to you to prove this is, in fact, a homomorphism.

Clearly, we have $\phi|_H = \text{id}_H$, and $\text{ker }\phi = N$.

However, $H \times N$ is abelian (since both $H,N$ are), while $G$ is not, so $G \neq H \times N$.

$\endgroup$
  • $\begingroup$ Thanks, for your answer. I am still going through it. Meanwhile, I have shown that $G=HN$, i.e., every element in G can be written as $hn$ where $h \in H$ and $h \in N$. So for every $g=hn \in G$ I have an element $(h,n)$ in $H \times N$. Isn't sufficient to prove $\psi: H \times N \to G$ surgective? $\endgroup$ – levitt Mar 29 '15 at 17:37
  • $\begingroup$ Your posted answer doesn't SHOW $G = HN$ it assumes it. I'm not saying that isn't a viable approach (in fact, it's true), however one CANNOT conclude that $G = HN$ and $H \cap N = \{e\}$ together imply $G \cong H \times N$, although one can conclude they are set-isomorphic. $\endgroup$ – David Wheeler Mar 29 '15 at 18:21
  • $\begingroup$ True, I didn't prove it here. I wanted to know if the approach is right to show the surgectivity. Thanks. I have used $G = HN$ to show surgectivity and then $H \cap N = \{e\}$ to show injectivity. After that homorphisim is also shown if $G$ is abelian. Aren't these sufficient for isomorphism? $\endgroup$ – levitt Mar 29 '15 at 18:29
  • $\begingroup$ Importantly, I want to know for sure, first of all, if I was right in concluding that $H \cap N = \{e\}$? $\endgroup$ – levitt Mar 29 '15 at 18:37
  • 1
    $\begingroup$ It is true: if $g \in \text{im }\phi \cap \text{ker }\phi = H \cap N$, then $\phi(g) = e$ and $\phi(g) = g$, so that $g = e$. $\endgroup$ – David Wheeler Mar 29 '15 at 18:43

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.