# The Unapologetic Mathematician

## Subgroups

A subgroup is pretty straightforward. It’s a little group living inside a bigger group. If you’ve got a group $G$ and some collection $H$ of elements of $G$ so that $H$ is a group using the same composition as $G$, then $H$ is a subgroup. To be more explicit, you need that

• If $x$ and $y$ are in $H$ then $xy$ is in $H$.
• If $x$ is in $H$ then $x^{-1}$ is in $H$.
• The identity $e$ of $G$ is in $H$. [added at the suggestion of Toby Bartels]

We say that a subgroup is “closed” under composition and inverse, meaning that if we start with elements of $H$ and take compositions and inverses we never leave the subgroup.

The collection of all even integers is a subgroup of the group ${\mathbb Z}$ of all integers (with addition as the operation).
The subset $\{e,(1\,2\,3),(1\,3\,2)\}$ is a subgroup of the group $S_3$.
Every group has two “trivial” subgroups: the whole group itself, and the subgroup consisting of just the identity element.
There are two ways of getting subgroups that I want to spend a bit more time on: “images” and “kernels”.

Let’s consider a homomorphism $f:G\rightarrow H$.

The image of $f$ — written ${\rm Im}(f)$ — is the collection of all elements of $H$ that actually occur as values of $f$. If $h_1=f(g_1)$ and $h_2=f(g_2)$ are such elements, then $h_1h_2=f(g_1)f(g_2)=f(g_1g_2)$, so it’s also in the image of $f$. Check for yourself that the image is closed under inverses and that it contains the identity of $H$. This shows that ${\rm Im}(f)$ is a subgroup of $H$.

The kernel of $f$ — written ${\rm Ker}(f)$ — is the collection of all elements of $G$ that are sent to the identity $e_H$ of $H$. If $g_1$ and $g_2$ are in the kernel, then $f(g_1g_2)=f(g_1)f(g_2)=e_He_H=e_H$, so $g_1g_2$ is also in the kernel of $f$. Again, verify for yourself that the kernel is also closed under inverses and contains the identity of $G$. This shows that ${\rm Ker}(f)$ is a subgroup of $G$.

Go back to last week’s post about homomorphisms and figure out the image and kernel of each of the examples. Also consider the following questions

• What is the kernel of a monomorphism?
• What is the image of an epimorphism?

February 13, 2007 -

1. Closure under composition and inverses is *not* enough to ensure that the identity belongs to the alleged subgroup. You also have to insist that the subgroup has at least one element. (The empty set is closed!) So really, it’s simplest just to stick that in as a third clause (that the subgroup must include the identity element).

At the very least, you have to say that a subgroup must be nonempty. I have seen a *lot* of books that accidentally leave this out, but I’ve never seen any reference that leaves it out *deliberately*. (That is, there doesn’t seem to be any disagreement about the correct definition, just forgetfulness.) Comment by Toby Bartels | February 15, 2007 | Reply

2. You know, I suppose I’ve never considered that point. It’s the sort of thing that once you know you don’t often think about, and which can trip you up when you try to teach it to someone else. Of course, I’ve not yet been allowed to teach anything but calculus…

Anyhow, the “subset that’s also a group with the same composition” definition is essentially what shows up in Judson and Hungerford, and I’ll bet most other introductory texts. In fact, Hungerford has an odd middle-ground that starts with “closed under product” and adds “is a group with the restricted composition” to that. Comment by John Armstrong | February 15, 2007 | Reply

3. Yeah, you can't go wrong with the ‘subset that's also a group with the same composition’ definition.

>Hungerford has an odd middle-ground that starts with “closed under product” and adds “is a group with the restricted composition” to that.

I'll bet that this is because they want the restricted composition to be well defined as a function from H × H to H. (But this isn't really necessary.)

[I hope that the HTML formatting shows up here!] Comment by Toby Bartels | March 5, 2007 | Reply

4. It works, but evidently something tripped the spam filter. Weird. Eerie. Comment by John Armstrong | March 7, 2007 | Reply

5. […] we know what a group is and what a subgroup is. Today I want to talk about the cosets of a subgroup in a group […]

Pingback by Cosets and quotients « The Unapologetic Mathematician | December 20, 2007 | Reply

6. […] of interesting representations that are also applicable to any group . The main ingredient is a subgroup — a subset of the elements of so that the inverse of any element in is also in , and the […]

Pingback by Coset Representations « The Unapologetic Mathematician | September 20, 2010 | Reply

7. […] not the kernel we’ve talked about recently, which is the kernel of a -morphism. This is the kernel of a group homomorphism. In this context, it’s the collection of group elements so that the image is the identity […]

Pingback by Lifting and Descending Representations « The Unapologetic Mathematician | October 29, 2010 | Reply