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Herstein defined the definition of a simple group as follows:

A group is said to be simple if it has no non-trivial homomorphic image.

Please help me to understand what is meant by non-trivial homomorphic image.

Thanks.

Cameron Buie
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Sriti Mallick
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    This is not the standard definition, according to which the groups of order 1 is not simple. – Derek Holt Apr 08 '13 at 17:11
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    The typical definition of simple group excludes the trivial (one element) group, likely so that we can assert all simple groups of odd order are actually of prime order. The definition quoted by OP does not have such an exclusion. – hardmath Apr 08 '13 at 18:40

5 Answers5

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Put another way, any homomorphism from the group is a monomorphism (so the image of the homomorphism is the group we started with, for all intents and purposes) or the homomorphism mapping everything to the identity (so the image of the group is the trivial subgroup generated by the identity of the group we're mapping to). If the homomorphism isn't injective and isn't the homomorphism mapping everything to the identity, then the group's image under the homomorphism doesn't "look like" the group, nor like the trivial subgroup. That's what Herstein means by "non-trivial homomorphic image."

Equivalently, the kernel of any homomorphism from the group is the trivial subgroup generated by the group's identity element, or it is the whole group. Since for any normal subgroup there is a homomorphism having it as a kernel (projection onto quotient group), and any homomorphism's kernel is a normal subgroup of the domain, then Herstein's definition is equivalent to: "A group is said to be simple if it has no non-trivial proper normal subgroups."

Cameron Buie
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  • It appears that it is in fact true that you can have a nontrivial non-injective surjective homomorphism $\varphi \colon G \to G'$ such that $G / \ker\varphi \cong G'$ under a different isomorphism. For example, the quotient of $\prod^\infty \mathbb{Z}$ by the normal subgroup $(\mathbb{Z}, 0, 0, \dots)$ is isomorphic to $\prod^\infty \mathbb{Z}$. Of course, this group certainly isn't simple... So, I don't know if it's actually better to think of "no nontrivial homomorphic image" to mean "all homomorphisms are either trivial or isomorphisms". – Doug Oct 14 '14 at 04:20
  • I'm actually just really confused now. – Doug Oct 14 '14 at 04:30
  • It seems to work perfectly when you interpret "no nontrivial homomorphic image" to mean "every surjective homomorphism is either trivial or an isomorphism". – Doug Oct 14 '14 at 04:40
  • That doesn't quite work. The trivial group is simple under Herstein's definition, yet under your definition, it wouldn't be, since every surjective homomorphism is both trivial and an isomorphism. The standard definition excludes the trivial group, though, so your definition is equivalent to that one. On the other hand, if you remove the word "either" from your definition, it becomes equivalent to Herstein's. – Cameron Buie Oct 14 '14 at 16:17
  • @CameronBuie I am still a bit confused as to why "...the group's image under the homomorphism doesn't "look like" the group, nor like the trivial subgroup". Doesn't Doug's example present a counterexample to this? There we have a nontrivial proper normal subgroup $N$ whose quotient group $G/N$ is isomorphic to $G$, no? – twosigma Sep 11 '20 at 15:35
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    @twosigma: Doug's example $G=\prod^\infty\Bbb Z$ is a non-simple group, because it has non-trivial homomorphic images. Consider the map $G\to\Bbb Z$ given by $$(n_1,n_2,n_3,\dots)\mapsto n_1.$$ – Cameron Buie Sep 11 '20 at 20:35
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There is a more down-to-earth definition:

A group $G$ is simple if $G$ has no nontrivial normal subgroup.

This definition better explains why simple groups are called simple, because containing no normal subgroups they cannot be broken up further.

Let's first see why the two definitions are equivalent. If $G$ has a normal subgroup $N$, then we have a group homomorphism \begin{equation} G\xrightarrow{\pi}G/N \end{equation} where $G/N$ is a homomorphic image of $G$, and is nontrivial (since $N$ is nontrivial).

If $G\xrightarrow{\phi}H$ is a group homomorphism where the image $\phi(G)\neq 0$ and $\phi(G)\neq G$, then $\operatorname{ker}(\phi)$ is a nontrivial normal subgroup. So the two definitions are indeed equivalent.

Although the two definitions are equivalent, they demonstrate two different view points. My down-to-earth one defines simplicity as a property of a group itself (in terms of certain subsets), while your definition defines this in terms of how $G$ interacts with other groups (in this case, it only interacts with $0$ and itself).

This view-point-shift seems to me a very important trend in modern mathematics. One very trivial yet illuminating example is the definition of injections and surjections.

So let \begin{equation} A\xrightarrow{f}B \end{equation} be a function between sets. Down-to-earth, we might call $f$ injective if $f(a)=f(a')$ implies $a=a'$. Again this defines injectivity as a property of the function itself. However we can also define this by looking at how $f$ interacts with other functions: $f$ is injective if \begin{equation} f\circ g_1=f\circ g_2 \end{equation} implies \begin{equation} g_1=g_2. \end{equation}It is not difficult to see these two definitions are equivalent but note that the second definition does not talk about maps or points or images. It characterizes injectivity in terms of how $f$ interacts with other functions, and it has a better chance to be generalized to other objects since we only need these objects to interact with each other (no need to have points or images).

There is a similar definition for surjections: $f$ is surjective if \begin{equation} g_1\circ f=g_2\circ f \end{equation} implies \begin{equation} g_1=g_2 \end{equation}.

Hui Yu
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An equivalent statement is to say that the group has no nontrivial normal subgroups. This is because if it had a nontrivial normal subgroup, that normal subgroup would be the kernel of some homomorphism, a contradiction to your definition.

Clayton
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The most important thing to realize is that homomorphic images are in correspondence with normal subgroups: given an homomorphic image of a group, you have an associated kernel. And given a normal subgroup, you have an associated homomorphic image of the group (the projection of the group onto its quotient).

Alex Provost
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The word trivial is used in mathematics in very context-dependent ways. Here non-trivial homomorphic image is understood to exclude two trivial cases that occur for every group.

A group $G$ has itself (or an isomorphic copy) as a homomorphic image, and also the image of a trivial group containing only one element (the trivial identity group). These two homomorphic images are considered trivial cases, and the definition of a simple group amounts to saying there are no other possibilities for a homomorphic image.

As other Answers point out, the definition of simple group is often stated as an equivalent property on normal subgroups, i.e. that there are only the group $G$ itself and the trivial (identity) subgroup which are normal in $G$. These forms of definition are equivalent by the First Isomorphism Theorem (for groups).

hardmath
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