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


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:

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


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 $G \neq H \times N$.

  • $\begingroup$ 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

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