Decomposition of a group

Asked Jan 21 '21 at 19:32

Active Jan 21 '21 at 20:06

Viewed 19 times

I have a small question about simple groups:

When we're decomposing a group $G$ we take a non-trivial normal subgroup of maximum order $G_1$ so that $\frac{G}{G_1}$ is simple and so on.

But why is $\frac{G}{G_1}$ simple?

I thought it had to do with the fact that $G_1$ is of maximum order, so there is no 'bigger' normal subgroup in $G$?

edited Jan 21 '21 at 20:06

Shaun

asked Jan 21 '21 at 19:32

Geigercounter

Yes, that's correct. Use the isomorphism theorems. – Qiaochu Yuan Jan 21 '21 at 19:37
@QiaochuYuan okay, how would I go about doing that? – Geigercounter Jan 21 '21 at 19:38
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Suppose $G/G_1$ isn't simple. Pick a nontrivial normal subgroup $N$ of it. Now consider the preimage of $N$ in $G$. What can you say about it? – Qiaochu Yuan Jan 21 '21 at 19:39
@QiaochuYuan I am kind of confused with what you are getting at... If we take a normal subgroup of $G/G_1$, how can we talk about it in $G$? – Geigercounter Jan 21 '21 at 19:44
By taking the preimage. If $N$ is a normal subgroup of $G/G_1$ then we can consider ${ g \in G : g + G_1 \in N }$. This is, by the isomorphism theorems, a normal subgroup in $G$, containing $N$. – Qiaochu Yuan Jan 21 '21 at 19:45
@QiaochuYuan then $G$ would have a normal subgroup with a higher order than $G_1$ thus we reach a contraposition? – Geigercounter Jan 21 '21 at 19:50
Perhaps also this duplicate is useful, mentioning the Correspondence Theorem. – Dietrich Burde Jan 21 '21 at 19:50
@DietrichBurde we haven't covered that yet in class I believe, but thanks for the link anyways! – Geigercounter Jan 21 '21 at 19:52
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You can see this directly, no problem. You see that several other people here had the same question. I think there are many nice answers given here. Have a look at them. – Dietrich Burde Jan 21 '21 at 19:54

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