Non-isomorphic groups

Dear Dr. Kim,

I am having a few problems with regards to some work on groups. I would like to know how to find isomorphically distinct groups in the context of a semi-direct product, given a number of possible homomorphisms.

Is it enough to show that if the center’s of each possible group are distinct?

If so, how would you find the center of semidirect product with C_10 x C_4 as the underlying set with h:C_4 –>Aut(C_10) a homomorphism, such that h(y) = phi_9, y belongs to C_4, phi_9 belongs to Aut(C_10).

I am aware that the centre is the fixed point set acting on itself by conjugation, however when I try to evaluate this set, denoted Z(g), I can only find that 1, x^5 belong to Z(g) where x^5 belongs to C_10. I have seen elsewhere that the center of this set is isomorphic to C_2 x C_2 but am unable to compute this. Do you see what I have overlooked?

Is this the most efficient way to find distinct groups and eliminate those that are not? I have seen some examples where two groups where found to be the same under a change of variable. Are there any rules or guidelines to take into account when looking for these things, or is it just a matter of remembering them?

Thanks very much for your time.

Reply:

I will answer the last question first: No mathematical problem is solved only by remembering a number of facts. Of course there are principles. Most importantly, make sure you understand the definitions. Your characterization of the center Z(g) is rather poorly phrased, but also rather too sophisticated in most concrete situations (although it may be useful in certain general theorems like the Sylow theorem). You should rather make sure you know the simple definition: Z(g) is the subset of g consisting of elements that commute with all elements of g. In symbols,

Z(g)=\{x\in g: \ \ \forall y\in g,\ \  xy=yx\}

Try to use this to compute the center of the group you describe above. In this regard, it is convenient to note that the group is generated by the elements (1,0) and (0,1). So an element x=(m,n) is in the center if and only if (m,n)(1,0)=(1,0)(m,n) and (m,n)(0,1)=(0,1)(m,n). If you just use the definition of the semi-direct product carefully, the rest should be easy.

Here are some general remarks in relation to the problem: To check if G_1 and G_2 are isomorphic, you should go through some features of groups that are preserved under isomorphisms. Typically, you start with the most simple-minded features and progress to more sophisticated ones. For example, if I ask you if C_5 and C_7 are isomorphic, you should be able to say right away that they are not, simply because they have different numbers of elements. That is, the cardinality of a group is preserved under isomorphism. How about S_3 and C_6? They are not isomorphic because one is commutative and the other is not. Again, the point is that the property of being commutative is preserved under isomorphism. After this, things can become considerably more complicated, depending on the example. But the criterion of computing the center is a good one: if G_1 and G_2 are isomorphic, then Z(G_1) and Z(G_2) are isomorphic. Therefore, if Z(G_1) and Z(G_2) are *not* isomorphic, then G_1 and G_2 are not isomorphic. Of course, this is useful only if it’s easier to distinguish Z(G_1) and Z(G_2) than to distinguish G_1 and G_2 directly. But it’s plausible that there is an `inductive’ structure to the problem, in that if the groups involved are non-abelian, then the centers are smaller than the original groups and furthermore, abelian, and therefore, easier to analyze.

The general problem of distinguishing groups can eventually involve a great deal of ingenuity and knowledge. For example, if Z(G_1) and Z(G_2) are isomorphic, it can be quite complicated. One good way to get a feel for such problems is via constructing yourself some tricky examples. Can you come up with G_1 and G_2 that are definitely non-isomorphic, but for which |G_1|=|G_2| and Z(G_1)\simeq Z(G_2)?

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