# Some intuitive vision on abstract algebra

I'm currently on my second year of Mathematics and first one I'm learning abstract algebra. So far it's being the most interesting subject I've seen, but I have a problem with some intuitive visions about it that some teachers (out of experience I guess) have with this.

Particularly, I heard one of them saying it was trivial that, under isomorphisms:

$\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z} \subset A_4$

That meaning a subgroup of $A_4$ is isomorphic to the left side of the expression. I don't see that trivial at all until you do some study on $A_4$ (btw, it's the alternating group of order 12), and you find that there are elements of order 2 x,y, for which is true that $xy=yx$, you can deduce the next is a subgroup of $A_4$: $H=\{1,x,y,xy\}$. Once yo get here, the isomorphism between $H$ and $\mathbb{Z}_2\times \mathbb{Z}_2$ may look trivial, but I don't understand how, from just an alternating group, you can deduce that just by looking.

Is there any intuitive vision on groups, either general or in this particular case, that I'm missing and could help me "see" these things?

Also, would you recommend some book that makes a special effort explaining, not the formalism, which you can find in any book, but this more intuitive view?

EDIT: Ok, so my first question has been answered, thank you all guys. Now, my second, about the book, if anyone knows a book that's "famous" for begin great in group theory (elementary group theory), or that deals with the subject at an intuitive level (I won't give up the formal treatment), I would appreciate a name or author.

EDIT2: @jspecter This was very usefull, thank you. A couple of questions, I don't understand the group $Stab_\Sigma(P)$: I'm not english an I don't know the meaning of "tabalizes", may be a typo... The second, I think I'm having some trouble thinking about the $\mathbb{Z}/m\times ... \times \mathbb{Z}/r$, I'm seeing them as ordered sets of as much integers as factors you have $(x_1,x_2,...,x_n)$, (I don't know the word for that in english either, we called them n-tuplas (spanish)). Maybe my difficulties are with seeing the isomorphism between these groups and the symmetry groups $A_n$. With this particular example, if we take your subset $P$, which we can see as a triangle in a plane, and we create the group of symmetries of this figure we have the identity and 3 reflexions on the mediatrice of each vertex, this is what we identify with $\mathbb{Z}_2\times \mathbb{Z}_2=\{(0,0),(1,0),(0,1),(1,1)\}$. Obviously we identify two reflexions with (1,0) and (0,1), and the other is a combination of the other two, plus it's abelian (just like any four element group), is that the way you're seeing it? I think I have difficulties identifying these members (i,j) with functions, in this case isometries for that triangle, any advice for this?

Overall, you helped a lot and I'm starting to see the whole thing more clear. Thank you again.

Still, if someone recommends a book, I would be gratefull.

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$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\cong\{id, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)\}\leq A_4$ – user1729 Jan 8 '13 at 15:31
(Find all the possible elements of order two in $S_4$. Which are in $A_4$? What subgroup do they form?) – user1729 Jan 8 '13 at 15:31
@user1729 Small typo: you missed out the identity permutation. I would still personally claim that this is trivial in the sense of being elementary, rather than in the sense of being obvious. – Matthew Pressland Jan 8 '13 at 15:33
@Ferfer93: The set $H=\{id, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)\}$ is a subgroup of $A_4$, by the definition of $A_4$. $H$ has order $4$, as it has $4$ elements. There are only two subgroups of order $4$ up to isomorphism. What are they, and why is $H\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$? – user1729 Jan 8 '13 at 15:36
Another, more geometric, perspective is that $A_4$ is the group of proper (rotation) symmetries of a regular tetrahedron. The subgroup isomorphic to $\mathbb{Z} / 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$ consists of the three "flips" (rotations which fix two opposite edges) and the identity. – Michael Joyce Jan 8 '13 at 15:53

One way to think about groups (or rather the only way) is as the symmetries of other objects. So if you want to 'see' facts about a group one way is to present it as the symmetries of something geometric.

Now we need some easily provable (but maybe non-intuitive) imput of a geometric object with $A_4$ - symmetry. One (mathematically precise) example is the regular tetrahedron $T \subset \mathbb{R}_3$ together with the subgroup $\Sigma$ of distance and orientation preserving maps from $\mathbb{R}^3$ to $\mathbb{R}^3$ which send $T$ to $T.$ But you should think of this as your favorite four sided die together with all the ways you could pick it up and put it back down on a table so that if you erased the numbers you wouldn't have any proof you moved it at all. Oh here's one now:

How might one see the symmetry group of this object is $A_4?$ Well, any symmetry of $T$ must send vertices to vertices. Thinking of the symmetries $\Sigma$ as the symmetries of the vertex set we obtain a map (homomorphism) from $\Sigma$ to $S_4.$ And because any symmetry which fixes all these vertices must be the identity this must map be an embedding. Checking $|\Sigma| = 12,$ it follows $\Sigma \cong A_4.$

Now if one wants to cut out a subgroup of a symmetry group of some object $O$ one way to do it is to look at all the symmetries preserving some subobject $O'.$

Consider the pair of edges consisting of the edge $l_1$ (in the front) where you can see the two adjacent faces and the edge $l_2$ (in the back) which is obscured by the placement of the die. Denote the subobject of $T$ consisting of the pair of marked lines $l_1$ and $l_2$ by $P.$

Forgetting that $P$ is a object of $T,$ the set $P$ has a group of distance and orientation preserving symmetries in it's own right. We denote this group by $\Sigma_P.$ That is $\Sigma_P$ consists of all distence and orientation preserving maps from $R^3 \rightarrow \mathbb{R}^3$ that take $l_1$ to $l_1$ and $l_2$ to $l_2.$ It's easy to see $\Sigma_P \cong \mathbb{Z}/2 \times \mathbb{Z}/2.$

There is also the subgroup of $\Sigma$ which stabilizes $P.$ We denote this subgroup $Stab_{\Sigma}(P).$ Since every element of $Stab_{\Sigma}(P)$ restricts to a distance and orientation preserving we obtain a restriction map

$$Stab_{\Sigma}(P) \rightarrow \Sigma_P$$

One now observes two things. Any symmetry of $P$ lifts to a symmetry of $T$ Any symmetry in $\Sigma$ which fixes $P$ pointwise is the identity (it fixes the vertices of T). From the former we obtain the above map is surjective and by the latter that the map is injective. And so

$$\mathbb{Z}/2 \times \mathbb{Z}/2 \cong \Sigma_P \cong Stab_{\Sigma}(P) \subset \Sigma \cong A_4$$

This argument is long winded. This seems to be how it is when you describe something geometric.

A picture costs 1000 words.

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Thank you for the answer. I edited the first post to ask you something, because the comment didn't look big enough. – MyUserIsThis Jan 8 '13 at 17:23

In this particular case you could argue like this to avoid computations. By Sylow's theorem $A_4$ has a subgroup of order $4$. It cannot be cyclic because there is no element of order $4$ in $A_4$. The only noncyclic group of order $4$ is $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, so the statement follows.

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(The downvote was me - I am sorry, but I do not think there is anything intuitive about Sylow's theorems! Nor does it avoid complications...) – user1729 Jan 8 '13 at 15:47
I agree that Sylow's theorems are not intuitive, after using them a while they become like second nature though. This is just a quick way to see the result using almost none of the structure of $A_4$ (really just using the fact that $A_4$ has order divisible by $4$ and no elements of order $4$), I was mostly going with "but I don't understand how, from just an alternating group, you can deduce that just by looking" from the question – Mikko Korhonen Jan 8 '13 at 17:06
Is there any intuitive vision on groups,
either general or in this particular case,
that I'm missing and could help me "see" these things?


Groups measure symmetry. We can understand groups by their groups actions, which are symmetries of a set (usually with additional structure).

Informally speaking, the set of all invertible maps from some mathematical object to itself which preserve some property will form a group. In fact, every group can be expressed in this way (via a rather formal-looking definition using category theory). In other words, every group is the automorphism group of some mathematical object.

This description of groups might not be obvious for every group you have seen so far, but do you see how symmetric groups fit this explanation? Dihedral groups? Matrix groups? What sets are they the symmetries of? What type of structure do we require these symmetries to preserve?

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This I hadn't heard about, so every group answers to some kind of symmetry of a set? Could you better define symmetry in this case? In the case of Dihedral or Symmetric groups it's pretty ovbious. In the case of Matrix groups, I've always seen them as groups of linear applications between vector spaces, that preserve that structure of a vector space, what kind of symmetry are you talking about in this case? – MyUserIsThis Jan 8 '13 at 16:25
@Ferfer93: A group is the set of symmetries of its Cayley graph. – user1729 Jan 8 '13 at 16:30
@user1729 Thanks, I'm discovering lots of things here... – MyUserIsThis Jan 8 '13 at 16:32
@Ferfer93 Yes, matrices preserve the linear structure, ie. addition and scalar multiplication. I was deliberately vague about what structure group operations preserve, because the set of all invertible maps from an object to itself preserving some specified structure will form a group, regardless of what that structure is. – Brett Frankel Jan 8 '13 at 19:10