## Review of Group Theory: Conjugacy Classes of The Product of Two Groups

**Point of Post: **In this post we show how one can completely classify the conjugacy classes of the product of two groups and given the conjugacy classes of each.

*Motivation*

It’s a natural question whether given two groups and and their set of conjugacy classes can we tell anything about the conjugacy classes of the product of the two groups. In fact, (as may or may not be obvious) there is an extremely easy way to find the set of all conjugacy classes of . Namely, if denotes the set of conjugacy classes of and those of then is equal to . Thus, in particular .

*Classifying the Conjugacy Classes of The Product Group*

We now aim to classify the conjugacy classes of given those of and . Indeed:

**Theorem: ***Let and be groups which have the sets of conjugacy classes and respectively. Then, if denotes the set of conjugacy classes of then*

* *

* *

**Proof: **Let and with and , we claim that . Indeed, let then there exists some such that and so and so that . Conversely, let then and for some and and so and so . Thus as claimed and so .

Conversely, let and let and where is the canonical projection onto the coordinate group. Our first claim is that . Indeed, evidently , so to show the reverse inclusion let . We evidently have then that there is some and so that and . Thus, by definition there must exists so that and similarly and so in particular and and so and so as desired. What we now claim then is that and . Indeed, let then and so for some and so in particular so that . If then since . Thus, , and similarly one can prove that . Thus, as required.

The conclusion follows.

**References:**

1. Rotman, Joseph J. *Advanced Modern Algebra*. Providence, RI: American Mathematical Society, 2010. Print.

[…] we’ve defined a map by which by the second part of our theorem is an injection. But, by a previous theorem we have that and thus by a simple set theoretic fact we may conclude that is also a surjection […]

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